GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Oct 2019, 20:25 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  After multiplying a positive integer A, which has n digits,

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager  Joined: 09 Feb 2013
Posts: 111
After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

5
1
34 00:00

Difficulty:   95% (hard)

Question Stats: 33% (02:18) correct 67% (02:20) wrong based on 233 sessions

HideShow timer Statistics

After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

_________________
Kudos will encourage many others, like me.
Good Questions also deserve few KUDOS.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9695
Location: Pune, India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

23
4
emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
General Discussion
MBA Section Director V
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 7083
City: Pune
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

None

1

2

8

9

Constraint 1) when we put a sequence of multiples of (n+2) atleast one multiple should have its unit digit same as that of (n+1)
Constraint 2) N+1 can not greater than 9 since it is a single digit.
Constraint 2) N can not be 0

n(n+1)(n+2)
8---9----10------ Units digit zero always. so out
7---8----9-------- 9X2=18 so 2323232X9 = .......2988 or 2222222 x 9 = ......98 out
6---7----8-------- units digits 8,6,4,2,4,8,6,4,2,0,8,6.... No 7 so out
5---6----7-------- 7x8=56 so 83838 x 7 = .......5866 or 88888 x 7 = ....216 out
4---5----6-------- units digits 6,2,8,4,0,6,2,8,4,0...... No 5 so out
3---4----5-------- 5,0,5,0 out
2---3----4-------- 4,8,2,6,0,4.... out
1---2----3-------- only possible pair is 3 x 4 = 12 so out

Bunuel, Can you Pls help?
_________________
2020 MBA Applicants: Introduce Yourself Here!

MBA Video Series - Video answers to specific components and questions about MBA applications.

2020 MBA Deadlines, Essay Questions and Analysis of all top MBA programs
MBA Section Director V
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 7083
City: Pune
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

VeritasPrepKarishma wrote:
emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

You said correctly Karishma.
It is important to take first step in right direction.

Regards,

Abhijit
_________________
2020 MBA Applicants: Introduce Yourself Here!

MBA Video Series - Video answers to specific components and questions about MBA applications.

2020 MBA Deadlines, Essay Questions and Analysis of all top MBA programs
Intern  Joined: 17 Aug 2015
Posts: 6
Location: India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

Hi Karishma,

How did you reach to 666666 specifically?

Also, the qualifying numbers in the problem could be
11,22,33,44,55,66,77,88,99
111,222,333,444,555,666,777,888,999
1111....
11111....
111111...
and on

How do we narrow down this in a short span of time? Sorry, I could not understand this.

Thanks,
Vikram
Director  P
Joined: 20 Sep 2016
Posts: 633
Location: India
Concentration: Strategy, Operations
GPA: 3.95
WE: Operations (Real Estate)
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

1
vikgupta07 wrote:
emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

Hi Karishma,

How did you reach to 666666 specifically?

Also, the qualifying numbers in the problem could be
11,22,33,44,55,66,77,88,99
111,222,333,444,555,666,777,888,999
1111....
11111....
111111...
and on

How do we narrow down this in a short span of time? Sorry, I could not understand this.

Thanks,
Vikram

After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

Read the question carefully:

You say 11 could be one of the numbers- here's how it cannot-

11 is the product of A with n number of digits and (n+2).
Now to get 11 we have to have number of digits to be 0. Only then can the number be (n+1=0+1) and as the question tells us that all the numbers are all same then in no way can you get 11 a 2 digit number of n =0 as the maximum the product could be is of 1 digit.

Posted from my mobile device
Intern  B
Joined: 30 May 2017
Posts: 19
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

Hello,
I do not understand the question and the answers.
I cannot see written in the text that all digits of A are the same.

let's say that A = 4,5,6,7, 8 or 9 in this case n=1. If I multiply A by 3 (n+2) I will have a number with 2 digits.

If I choose a number with two digits: A= 25, 26, 27,..., 99 then A*4 will have 3 digits and so on..

So what didn't I understand?
Director  P
Joined: 20 Sep 2016
Posts: 633
Location: India
Concentration: Strategy, Operations
GPA: 3.95
WE: Operations (Real Estate)
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

amascarenhas wrote:
Hello,
I do not understand the question and the answers.
I cannot see written in the text that all digits of A are the same.

let's say that A = 4,5,6,7, 8 or 9 in this case n=1. If I multiply A by 3 (n+2) I will have a number with 2 digits.

If I choose a number with two digits: A= 25, 26, 27,..., 99 then A*4 will have 3 digits and so on..

So what didn't I understand?

You may have been right about not understanding the question.

You got the first 2 parts right.

Third part- we get a number with (n+1) digits, all of whose digits are (n+1).

This means that product of A and n+2 has every digit same.
As in A= 95238 >>number of digits (n)= 5

Therefore A*(5+2) = 666666

The answer has 6 ie.(n+1) digits . And each digit in the answer is of the form (n+1)

So basically if you consider
Say A=23 >> n=2
A* (n+2) = (n+1)*(n+1)*(n+1)

Here the answer has three digits ie. (n+1) digit with each digit as (n+1)

Now the above example actually does not exist as there is only one such number and that is 666666

Posted from my mobile device
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9695
Location: Pune, India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

amascarenhas wrote:
Hello,
I do not understand the question and the answers.
I cannot see written in the text that all digits of A are the same.

let's say that A = 4,5,6,7, 8 or 9 in this case n=1. If I multiply A by 3 (n+2) I will have a number with 2 digits.

If I choose a number with two digits: A= 25, 26, 27,..., 99 then A*4 will have 3 digits and so on..

So what didn't I understand?

Read this: "... we get a number with (n+1) digits, all of whose digits are (n+1)"

The number has (n+1) digits. Each digit is (n+1).
So the possible numbers are:
1 digit number with each digit 1 i.e. 1
2 digit number with each digit 2 i.e. 22
3 digit number with each digit 3 i.e. 333
and so on...
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  B
Joined: 25 Jul 2012
Posts: 23
Location: India
Concentration: Finance, Marketing
GMAT 1: 660 Q44 V37 Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

Bunuel - Can you please help with this one. Thanks
_________________
If you like my post, consider giving me KUDOS Intern  Joined: 17 Aug 2015
Posts: 6
Location: India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

Read the question carefully:

You say 11 could be one of the numbers- here's how it cannot-

11 is the product of A with n number of digits and (n+2).
Now to get 11 we have to have number of digits to be 0. Only then can the number be (n+1=0+1) and as the question tells us that all the numbers are all same then in no way can you get 11 a 2 digit number of n =0 as the maximum the product could be is of 1 digit.

Posted from my mobile device[/quote]

Thank you!

Your comment did somehow helped me solve my doubt. I got it sorted out.
Intern  Joined: 17 Aug 2015
Posts: 6
Location: India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

780gmatpossible wrote:
emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

Bunuel - Can you please help with this one. Thanks

Hi,

Let me try and explain

1. A=Positive Integer [N Digits], Let's say 5
2. Multiplication Factor: [N+2], From 1, it's 5+2=7
3. Resultant Number: 5*7=35, which has [N+1] digits, according to the question, but all digits are not [N+1], The required number, according to this case should have been 66

The only numbers that would satisfy the problem are: -

1
22
333
4444
55555
666666
7777777

and so on.

If this is the resultant number, then if it has [N] digits, it must have been divisible by [N+1], FROM 2 Above. Do the quick math as to what number is divisible by [N+1]

from the qualifying set above, 666666/7=95238 is a candidate.

This A should satisfy all boundaries of the problem. therefore instances=1 [B ]

Hope this helps.

Thanks,
VIKRAM
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

$$\left. \matrix{ A = \,\,\underbrace {\underline {} \,\,\underline {} \,\,\underline {} \,\, \ldots \,\,\underline {} }_{n\,\,{\rm{digits}}}\,\,\,\, \ge 1\,\,{\mathop{\rm int}} \hfill \cr A \cdot \left( {n + 2} \right) = \underbrace {\underline {n + 1} \,\,\underline {n + 1} \,\,\underline {n + 1} \,\, \ldots \,\,\underline {n + 1} }_{n + 1\,\,{\rm{digits}}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,?\,\,\,\, = \,\,\,\,\# \,\,A\,\,\,{\rm{possible}}$$

This is a typical organized manual work technique exercise!

$$\left\{ \matrix{ n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\, \cdot \,\,\,\left( {1 + 2} \right) = \underline 2 \,\,\underline 2 \,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{22} \over 3} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 2\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {2 + 2} \right) = \underline 3 \,\,\underline 3 \,\,\underline 3 \,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{{\rm{odd}}} \over {{\rm{even}}}} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 3\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {3 + 2} \right) = \underline 4 \,\,\underline 4 \,\,\underline 4 \,\,\underline 4 \,\,\,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{4444} \over 5} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 4\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr n = 5\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {5 + 2} \right) = \underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\,\,\,\,\,,\,\,\,\,\underline {{\rm{viable}}} \,\,{\rm{solution}}\,\,\,\,\,\left( {{{666666} \over 7} = 95238} \right) \hfill \cr n = 6\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr n = 7\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {7 + 2} \right) = \underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{88888888} \over 9} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 8\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr n = 9\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {9 + 2} \right) = \underline {10} \,\,\underline {10} \,\,\underline {10} \,\, \ldots \,\,\underline {10} \,\,\,????\,\,\,\,{\rm{impossible}}\, \hfill \cr n \ge 10\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 9} \right) \hfill \cr} \right.$$

The correct answer is therefore (B). (This is all VERY fast, although hard to type!)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8040
Location: United States (CA)
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

emmak wrote:
After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

Let’s test some values for n.

If n = 1, A is a 1-digit number. We multiply A by 3 and we get a 2-digit number, which is 22. However, 22 is not a multiple of 3. So n can’t be 1.

If n = 2, A is a 2-digit number. We multiply A by 4 and we get a 3-digit number, which is 333. However, 333 is not a multiple of 4. So n can’t be 2.

If n = 3, A is a 3-digit number. We multiply A by 5 and we get a 4-digit number, which is 4444. However, 4444 is not a multiple of 5. So n can’t be 3.

If n = 4, A is a 4-digit number. We multiply A by 6 and we get a 5-digit number, which is 55,555. However, 55,555 is not a multiple of 6. So n can’t be 4.

If n = 5, A is a 5-digit number. We multiply A by 7 and we get a 6-digit number, which is 666,666. We see that 666,666 is a multiple of 7 (666,666 = 7 x 95,238)! So n can be 5.

At this point, we can skip even values of n, since the (n+1)-digit number it forms is odd and will never be a number of n + 2, which is even.

If n = 7, A is a 7-digit number. We multiply A by 9 and we get a 6-digit number, which is 88,888,888. However, 88,888,888 is not a multiple of 9. So n can’t be 7.

If n = 9, A is a 9-digit number. We multiply A by 11 and we get a 10-digit number. However, we can’t have a 10-digit number in which each of its digits is 10. Therefore, n can’t be 9, and we can stop here.

There is only one instance, n = 5, where all the criteria are met.

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15227
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: After multiplying a positive integer A, which has n digits,  [#permalink]

Show Tags

Hi Gmat_mission,

We're told that after multiplying a positive integer A (which has N digits) by (N+2), we get a number with (N+1) digits, all of whose DIGITS are (N+1). We're asked for the number of possible values of A that 'fit' this description.

To start, this question certainly 'feels' weird - and you would likely find it easiest to 'work back' from the later pieces of information that you're given - and use 'brute force' (along with some Number Properties) to find the solution. We're looking to end up with a number that has (N+1) digits - and ALL of those DIGITS equal (N+1). Since we're dealing with digits, there are only a limited number of possible values that we can end up with:

22
333
4444
55555
666666
7777777
88888888
999999999

Thus, there are no more than 8 possibilities; to get the correct answer, we have to incorporate the other pieces of information that we're given and see which of these end numbers actually 'fits' everything that we're told.
IF....
The end result was 22, then N=1, but there is no 1-digit number that you can multiply by (1+2) = 3 and end up with 22. This is NOT possible.
The end result was 333, then N=2, but there is no 2-digit number that you can multiply by (2+2) = 4 and end up an ODD. This is NOT possible.
The end result was 4444, then N=3, but there is no 3-digit number that you can multiply by (3+2) = 5 and end up with 4444. This is NOT possible.
The end result was 55555, then N=4, but there is no 4-digit number that you can multiply by (4+2) = 6 and end up with ODD. This is NOT possible.

The end result was 666666, then N=5... there IS a 5-digit number that you can multiply by (5+2) = 7 and end up with 666666 (it's 95,238 - you just have to do a little division to prove it). This IS a possibility.

The end result was 7777777, then N=6, but there is no 6-digit number that you can multiply by (6+2) = 8 and end up with ODD. This is NOT possible.
The end result was 88888888, then N=7, but there is no 7-digit number that you can multiply by (7+2) = 9 (since 88888888 is NOT a multiple of 9). This is NOT possible.
The end result was 999999999, then N=8, but there is no 8-digit number that you can multiply by (8+2) = 10 and end up with ODD. This is NOT possible.

Thus, there's just one answer that 'fits' everything that we're told.

GMAT assassins aren't born, they're made,
Rich
_________________ Re: After multiplying a positive integer A, which has n digits,   [#permalink] 15 Mar 2019, 15:25
Display posts from previous: Sort by

After multiplying a positive integer A, which has n digits,

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  