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# A number when divided successively by 4 and 5 leaves remainders 1 and

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A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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20 Oct 2015, 12:50
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Question Stats:

31% (01:11) correct 69% (01:34) wrong based on 488 sessions

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A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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20 Oct 2015, 21:48
15
4
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

Here, the concept is of successive division
i.e. the no is first divided by 4 and it leaves remainder 1 and quotient is let x,

therefore we have $$number = 4x + 1$$

and then the quotient x is divided by 5 and the remainder is 4

so, we have x in the form of $$x = 5k + 4$$

putting this value of x above, we get

$$number = 4(5k + 4) +1$$

$$=> number = 20k + 17$$

when this number will be divided by 20, we will get remainder = 17

Kudos, if you like the explanation
##### General Discussion
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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20 Oct 2015, 13:00
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?[/b]

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

t=4a+1 and t=5b+4=4b+4+b
=> b=4m+1
=> t=20m+9
=> the remainder is 9 when t is divided by 9

Ans: D
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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20 Oct 2015, 15:37
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

When it comes to remainders, we have a nice rule that says:

If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

A number when divided successively by 4 leaves a remainder 1
Possible values of the number are: 1, 5, 9, 13, 17, 21,...

A number when divided successively by 5 leaves a remainder 4
Possible values of the number are: 4, 9...STOP!

Both lists contain 9, so this could be the number.

What will be the remainder when this number is divided by 20?
9 divided by 20 = 0 with remainder 9

Cheers,
Brent
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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20 Oct 2015, 21:31
2
GMATPrepNow wrote:
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

When it comes to remainders, we have a nice rule that says:

If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

A number when divided successively by 4 leaves a remainder 1
Possible values of the number are: 1, 5, 9, 13, 17, 21,...

A number when divided successively by 5 leaves a remainder 4
Possible values of the number are: 4, 9...STOP!

Both lists contain 9, so this could be the number.

What will be the remainder when this number is divided by 20?
9 divided by 20 = 0 with remainder 9

Cheers,
Brent

I think the keyword here is that the number is divided successively .
Let x be the number which when successively divided by 4 and 5 gives 1 and 4 as remainder.

1. When we first divide the number by 4 .
x = 4y +1
where y is quotient from the first division

2. Now as the division is done successively , we divide the quotient from the first step by 5
y= 5z +4
where z is quotient from the second division
If Z=1 ,
y = 9
x= 9*4 + 1
= 37

Now , on dividing the by 20 , we get 17

Alternatively , we can also proceed from the final quotient .

Let x be the quotient after dividing the original number by 5.
So , the number which we divided by 5 to yield a remainder of 4 = 5x+ 4

This number 5x+4 is the quotient from the first division of the original number by 4
Therefore , the number should be = 4(5x+4) + 1
= 20x +17

We can clearly see that when divided by 20 , we should get a remainder of 17

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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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21 Oct 2015, 08:41
skywalker18 wrote:
I think the keyword here is that the number is divided successively .
Let x be the number which when successively divided by 4 and 5 gives 1 and 4 as remainder.

1. When we first divide the number by 4 .
x = 4y +1
where y is quotient from the first division

2. Now as the division is done successively , we divide the quotient from the first step by 5
y= 5z +4
where z is quotient from the second division
If Z=1 ,
y = 9
x= 9*4 + 1
= 37

Now , on dividing the by 20 , we get 17

Alternatively , we can also proceed from the final quotient .

Let x be the quotient after dividing the original number by 5.
So , the number which we divided by 5 to yield a remainder of 4 = 5x+ 4

This number 5x+4 is the quotient from the first division of the original number by 4
Therefore , the number should be = 4(5x+4) + 1
= 20x +17

We can clearly see that when divided by 20 , we should get a remainder of 17

I have to admit that, when I saw the word "successively," I thought that this might be what it means. However, I soon discounted that possibility and assumed the intent of the question was to tell us that, "When a certain number is divided by 4 and 5, the remainders are 1 and 4 respectively"

Here's why I discounted the other possibility:

In order for us to proceed as you have suggested above, we must ignore some important information. For example, let's divide 49 "successively" by 5 and 3. First, 49 divided by 5 equals 9 with remainder 4. This means that 49 divided by 5 equals 9 4/5. Now, we must take the result and divide by 3. You are suggesting that we take 9 and divide by 3. What about the the remainder of 4? What do we do with that? Alternatively, what do we do with the 4/5? We ignore it? What is there in the question that tells us to ignore the remainder from the first division?

We could also note that dividing a number successively by 5 and 3, is the same as dividing a number by 15. If we divide 49 by 15, we get a remainder of 4. If we apply the operations as you suggestabove, we get a remainder of 0.

The concept of successively dividing only seems to make sense (in my opinion) if the original number is divisible by BOTH divisors. For example, it makes sense to say that dividing 48 successively by 3 and 2 yields a result of 8. That is, 48/3 = 16 and 16/2 = 8.

Ignoring a certain component (i.e., the remainder) of the first division when performing the second division is not stated anywhere in the question. Given this, I don't think this could ever be an official GMAT question. I have certainly never seen a similar question. Have any other experts seen an official question of this nature?

Cheers,
Brent
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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21 Oct 2015, 14:14
4x+1= 1,5,9,13,17,21,25,29,33,37,41,45,49...
5x+4= 4,9,14,19,24,29,39,44,49....
29 & 49 are common in both so if we divide 29 or 49 by 20, we get 9

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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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21 Oct 2015, 14:46
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

Using cross multiplication formula

4 5

1 4

(4*4)+1 = 17

Good to learn this formula from Bunuel Thread only. It works great.

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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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21 Oct 2015, 21:33
GMATPrepNow wrote:
skywalker18 wrote:
I think the keyword here is that the number is divided successively .
Let x be the number which when successively divided by 4 and 5 gives 1 and 4 as remainder.

1. When we first divide the number by 4 .
x = 4y +1
where y is quotient from the first division

2. Now as the division is done successively , we divide the quotient from the first step by 5
y= 5z +4
where z is quotient from the second division
If Z=1 ,
y = 9
x= 9*4 + 1
= 37

Now , on dividing the by 20 , we get 17

Alternatively , we can also proceed from the final quotient .

Let x be the quotient after dividing the original number by 5.
So , the number which we divided by 5 to yield a remainder of 4 = 5x+ 4

This number 5x+4 is the quotient from the first division of the original number by 4
Therefore , the number should be = 4(5x+4) + 1
= 20x +17

We can clearly see that when divided by 20 , we should get a remainder of 17

I have to admit that, when I saw the word "successively," I thought that this might be what it means. However, I soon discounted that possibility and assumed the intent of the question was to tell us that, "When a certain number is divided by 4 and 5, the remainders are 1 and 4 respectively"

Here's why I discounted the other possibility:

In order for us to proceed as you have suggested above, we must ignore some important information. For example, let's divide 49 "successively" by 5 and 3. First, 49 divided by 5 equals 9 with remainder 4. This means that 49 divided by 5 equals 9 4/5. Now, we must take the result and divide by 3. You are suggesting that we take 9 and divide by 3. What about the the remainder of 4? What do we do with that? Alternatively, what do we do with the 4/5? We ignore it? What is there in the question that tells us to ignore the remainder from the first division?

We could also note that dividing a number successively by 5 and 3, is the same as dividing a number by 15. If we divide 49 by 15, we get a remainder of 4. If we apply the operations as you suggestabove, we get a remainder of 0.

The concept of successively dividing only seems to make sense (in my opinion) if the original number is divisible by BOTH divisors. For example, it makes sense to say that dividing 48 successively by 3 and 2 yields a result of 8. That is, 48/3 = 16 and 16/2 = 8.

Ignoring a certain component (i.e., the remainder) of the first division when performing the second division is not stated anywhere in the question. Given this, I don't think this could ever be an official GMAT question. I have certainly never seen a similar question. Have any other experts seen an official question of this nature?

Cheers,
Brent

Hi Brent,

If we take 37 as the original dividend and proceed with this question-
1. Dividing by 4
---------Q------------R
37/4 --- 9 --------- 1 -- 9*4 + 1 = 37
or ----- 9.25 ------ 0 -- 9.25*4 = 37
2. Dividing by 5
---------Q----------R
9/5 --- 1 --------- 4
or 9.25/5------ 1.85------ 0

3.On Dividing 37 by 20
---------Q----------R
37/20 --- 1----- 17
or ------1.85 ----- 0

In my opinion we can either express division -
1.As an integer quotient Q and a remainder R( which might be equal to 0)
2.Or as a decimal / fraction quotient Q, but in this case the remainder R will be always 0 .

But , we can't do both - have a decimal / fraction quotient and a remainder .
In general , we use method 1 and even here we are given a remainder .
If we use method 2 , we will get the same quotient after steps 1-2 and in step 3 , but the remainder will always be 0 .
The numbers used are similar to the one's suggested by you - dividing 49 successively by 5 and 3 or dividing by 15 ,
as 20 can be expressed as 5*4 .
Just my 2 cents .
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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21 Jun 2016, 21:19
Lets say the number N is divided successively by 4 & 5.

N= 4q+1

Successive Division,

q = 5 S + 4

Substituting,

N= 20 S + 17.

Hence, the remainder is 17 when divided by 20.
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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21 Jun 2016, 21:47
1
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

First it is important to understand successive division.

Given No. is divided by 4. Then whatever is the quotient, that quotient is divide by 5.

Let number is N.

N is of the type 4a+1 (Where a is any integer)

Now when N is divided by 4, we get "a" as the quotient.

so a is of the type: 5b+4

Ultimately N is of the type: 4*(5b+4) + 1

N = 20b+16+1 = 20b+17

when N is divided by 20, remainder will be 17.

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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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27 Oct 2016, 05:42
Again ..! Arghhh..

One of those Veritas prep Out of bound Questions.
Personal Opinion =>Veritas prep questions are either 500 level or out of scope of the gmat.
How on earth are we supposed to solve this question.
The only way someone can solve this up on the test day this if he/she has a PHD in mathematics.
Sigh.!

Abhishek009 How do i solve this one?
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A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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27 Oct 2016, 09:12
stonecold wrote:
Again ..! Arghhh..

One of those Veritas prep Out of bound Questions.
Personal Opinion =>Veritas prep questions are either 500 level or out of scope of the gmat.
How on earth are we supposed to solve this question.
The only way someone can solve this up on the test day this if he/she has a PHD in mathematics.
Sigh.!

Abhishek009 How do i solve this one?

Just posting my approach-

Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

No algebra required for solving this question since all the options are less than 20...

Try each option individually -

(A) 0/4 = 0 (Quotient) & 0(Remainder) further 0/5 = 0 (Quotient) & 0(Remainder)

(B) 3/4 = 3 (Remainder)

(C) 4/4 = 1 (Quotient) & 0(Remainder)

(D) 9/4 = 2 (Quotient) & 1 (Remainder) further 2/5 = 2 (Quotient) & 1(Remainder)

(E) 17/4 = 4 (Quotient) & 1 (Remainder) further 4/5 = 4 (Remainder)

Thus 17/20 will have remainder as 17, hence answer will be (E)...

For an algebraic approach find Here
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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27 Oct 2016, 09:17
1
Abhishek009 wrote:
stonecold wrote:
Again ..! Arghhh..

One of those Veritas prep Out of bound Questions.
Personal Opinion =>Veritas prep questions are either 500 level or out of scope of the gmat.
How on earth are we supposed to solve this question.
The only way someone can solve this up on the test day this if he/she has a PHD in mathematics.
Sigh.!

Abhishek009 How do i solve this one?

Just posting my approach-

Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

No algebra required for solving this question since all the options are less than 20...

Try each option individually -

(A) 0/4 = 0 (Quotient) & 0(Remainder) further 0/5 = 0 (Quotient) & 0(Remainder)

(B) 3/4 = 3 (Remainder)

(C) 4/4 = 1 (Quotient) & 0(Remainder)

(D) 9/4 = 2 (Quotient) & 1 (Remainder) further 2/5 = 2 (Quotient) & 1(Remainder)

(E) 17/4 = 4 (Quotient) & 1 (Remainder) further 4/5 = 4 (Remainder)

Thus 17/20 will have remainder as 17, hence answer will be (E)...

Hmm Back solving Works Here.
Great Job man

Thanks
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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07 Dec 2016, 17:54
N=4r+1
r=5s+4

Combine.

N=4(5s+4)+1 = 20s+16+1 = 20s+17 --> 20 doesn't go into 17, so the remainder is 17.

E.
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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27 May 2017, 09:27
2
hi experts , Please let me know if this is the right way to solve :

N/4= a(1) ------ (we can write this 4a+1=N) let N be any integer, a be the quotient and 1 is the remainder we get when we divide it by 4, as per question
now, a/5=b(4) ----- now the quotient 'a' is divided by 5 giving remainder 4.(let 'b' be the quotient in this case)
So, a/5=b(4) ----> we can write this as 5b+4 = a (possible value of a is : 4,9,14...)

lets put these values of a in equation 4a+1 = N ---> we get N = 17,37,57

All these values of N when divide by 20 ... gives us a remainder 17 hence answer is E.
Please explain if any shortcut is there to solve these kind of problem.
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A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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06 Jun 2017, 04:02
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

My way of solving this question was working with the answer choices, the number we are looking for should be there, isn't it ?
the answer choices represent the remainder when dividing by 20, so all we have to do in order to get the number is to add 20 and check if the number fits to the rules(working backwards):

(A) 0 + 20 = 20
according to the first rule, if we subtract 1 (the remainder) from the number, the result should be divided by 4
let's try:
20 - 1 = 19 -----> not divided by 4

next:
(B) 3 + 20 = 23
23 - 1 = 22 ------> not divided by 4

(C) 4 + 20 = 24
24 - 1 = 23 -----> not divided by 4

(D) 9 + 20 = 29
29 - 1 = 28 Great!, we can divide by 4. now we need to check for the second rule:

28 - 4 (the remainder of 2nd rule) = 24 ----> Since it's not divided by 5 it's not the answer

we're left with (E) then. let's check anyway
17 + 20 = 37
1st rule:
37 - 1 = 36 Great!, we can divide by 4

$$\frac{36}{4}$$ = 9

2nd rule:

9 - 4(remainder of the 2nd rule) = 5 Great, we can divide by 5, That's the answer

(E)

Give Kudos if like this approach
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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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08 Jun 2017, 16:45
2
1
Bunuel wrote:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

(A) 0
(B) 3
(C) 4
(D) 9
(E) 17

We need to find a number that when divided by 4 leaves a remainder of 1 and when the quotient from this division is divided by 5 leaves a remainder of 4. Let’s represent this number by n.

Since our number produces a remainder of 1 when divided by 4, it must be true that n = 4p + 1 for some integer p.

Since the quotient from the previous division, which is p, produces a remainder of 4 when divided by 5, we have p = 5q + 4. Let’s substitute this expression of p into the previous equation:

n = 4p + 1

n = 4(5q + 4) + 1

n = 20q + 16 + 1

n = 20q + 17

Finally, since 20q is divisible by 20, the remainder from the division of n by 20 is 17.

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Re: A number when divided successively by 4 and 5 leaves remainders 1 and  [#permalink]

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