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How many positive two-digit integers have a remainder of 1 when divide

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How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 04 Dec 2018, 00:57
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[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)

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How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post Updated on: 04 Dec 2018, 01:20
1
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,59 and 89

C is the answer.
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Originally posted by Chethan92 on 04 Dec 2018, 01:08.
Last edited by Chethan92 on 04 Dec 2018, 01:20, edited 1 time in total.
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How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 04 Dec 2018, 01:13
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)



took >2 mins to solve

first list out all 2 digit no which when divided by 5 would give remainder 4 : 14,19,24,29,34,39,44,49,54,59,64,69,74,79,84,89,94,99

all even no as they will be divisible by 2 and no divisible by 3 to be removed :
we are then left with 3 no. only :
29,59 & 89 IMO C ...
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 04 Dec 2018, 01:17
Afc0892 wrote:
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,58 and 87

C is the answer.

Afc0892
values would be 29,59 & 89

as per formula '30n-1'
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 04 Dec 2018, 01:21
Archit3110 wrote:
Afc0892 wrote:
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,58 and 87

C is the answer.

Afc0892
values would be 29,59 & 89

as per formula '30n-1'


Edited. Thank you.
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 04 Dec 2018, 05:05
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A quick lesson on remainders:

Quote:
When x is divided by 5, the remainder is 3.
In other words, x is 3 more than a multiple of 5:
x = 5a + 3.

When x is divided by 7, the remainder is 4.
In other words, x is 4 more than a multiple of 7:
x = 7b + 4.

Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.

To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.

Putting it all together:
x = 35c + 18.

Another example:
When x is divided by 3, the remainder is 1.
x = 3a + 1 = 1, 4, 7, 10, 13...
When x is divided by 11, the remainder is 2.
x = 11b + 2 = 2, 13...

Thus, when x is divided by 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...


Onto the problem at hand:

Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

A. 1
B. 2
C. 3
D. 4
E. 5


a remainder of 1 when divided by 2
x = 2a + 1 = 1, 3, 5, 7...
In other words, x must be ODD.

a remainder of 4 when divided by 5
x = 5b + 4 = 4, 9, 14, 19...
Since x must be odd, we get:
x = 9, 19, 29....

a remainder of 2 when divided by 3
In the blue list above, 29 is smallest value that yields a remainder of 2 when divided by 3.

Thus, when x is divided by 30 -- the LCM of the three divisors 2, 3, and 5 -- the remainder will be 29 (the smallest value that satisfies all of the given conditions):
x = 30c + 29 = 29, 59, 89, 119...

In the resulting list of options for x, only the 3 values in green are two-digit integers.


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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 04 Dec 2018, 18:55
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)

\(?\,\,:\,\,\# \,\,N\,\,{\rm{with}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right)\)

\(10\,\, \le \,\,N\,\,{\mathop{\rm int}} \,\, \le \,\,99\,\,\left( 1 \right)\)

\(\left\{ \matrix{
\,N = 2M + 1,\,\,M\,\,{\mathop{\rm int}} \,\,\,\,\left( {2{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( { - 3 \cdot 5} \right)} \,\,\,\,\,\,\, - 15N = - 30M - 15\,\,\,\,\left( {2{\rm{b}}} \right) \hfill \cr
\,N = 3K + 2,\,\,K\,\,{\mathop{\rm int}} \,\,\,\,\left( {3{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( {2 \cdot 5} \right)} \,\,\,\,\,\,\,10N = 30K + 20\,\,\,\,\left( {3{\rm{b}}} \right) \hfill \cr
\,N = 5L + 4,\,\,L\,\,{\mathop{\rm int}} \,\,\,\,\left( {4{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( {2 \cdot 3} \right)} \,\,\,\,\,\,\,6N = 30L + 24\,\,\,\,\left( {4{\rm{b}}} \right) \hfill \cr} \right.\)

\(\mathop \Rightarrow \limits^{{\rm{sum}}\,\,\left( {{\rm{2b}}{\rm{,3b}}{\rm{,4b}}} \right)} \,\,\,\,N = 30W + 29,\,\,W\,\,{\mathop{\rm int}} \,\,\,\,\,\left\{ \matrix{
\,W = 0\,\,\, \to \,\,\,N = 29\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr
\,W = 1\,\,\, \to \,\,\,N = 59\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr
\,W = 2\,\,\, \to \,\,\,N = 89\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr} \right.\)

\(? = 3\)


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

Regards,
Fabio.
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 06 Dec 2018, 00:30
=>

Let \(x\) be a positive integer with these properties.
Since \(x = 2p + 1\) for some integer \(p\), the possible values of \(x\) are \(x = 1, 3, 5, 7, … .\)
Since \(x = 3q + 2\) some integer \(q\), the possible values of \(x\) are \(x = 2, 5, 8, 11, … .\)
Since \(x = 5r + 4\) for some integer \(4\), the possible values of \(x\) are \(x = 4, 9, 14, 19, … .\)
The first possible 2-digit number is thus \(19\). To find the others, note that the least common multiple of \(2, 3\) and \(5\) is \(lcm(2,3,5) = 30.\)
Thus, there are three possible 2-digit numbers with these properties:
\(19, 49 = 19 + 30\) and \(79 = 49 + 30.\)

Therefore, C is the answer.
Answer: C
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How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 26 Aug 2019, 07:47
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)


Asked: How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)

N = 1 mod 2 = 2 mod 3 = 4 mod 5
N = 2k1 + 1 = 3k2 + 2 = 5k4 + 4
2k1 + 1 = 3k2 + 2 = odd = {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} = 5 + 6k5
5 + 6k5 = 5k4 + 4 = {29, 59, 89}

IMO C

Alternatively

N = -1 mod 2 = -1 mod 3 = -1 mod 5 = 2k1 -1 = 3k2 -1 = 5k3 -1
= LCM(2,3,5) -1 = 30k -1 = {29,59,89}

IMO C
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 26 Aug 2019, 08:25
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)


a1= 29

the said series is going to be of the form 30n-1.. so, only 3 numbers possible: 29, 59, 89.
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 23 Oct 2019, 00:05
azhrhasan wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)


a1= 29

the said series is going to be of the form 30n-1.. so, only 3 numbers possible: 29, 59, 89.


Hi how do you come up with 30n-1 ? I understand that 30 is the LCM but not quite get how to come up with 1 there.
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 23 Oct 2019, 18:32
David nguyen wrote:
azhrhasan wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(4\) when divided by \(5\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. 5\)


a1= 29

the said series is going to be of the form 30n-1.. so, only 3 numbers possible: 29, 59, 89.


Hi how do you come up with 30n-1 ? I understand that 30 is the LCM but not quite get how to come up with 1 there.


find the first number in the series and express it in terms of 30n + x
Here the first number in series is 29. therefore, 30n-1.

cool ?
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 23 Oct 2019, 18:54
Chethan92 wrote:
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,59 and 89

C is the answer.


can you kindly explain why did you take common difference as 1?
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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New post 23 Oct 2019, 20:29
pulak1988 wrote:
Chethan92 wrote:
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,59 and 89

C is the answer.


can you kindly explain why did you take common difference as 1?


Have a look at the remainders that you get when a number is divided by 2,3 and 5..

A number divided by 2 gives you a remainder of 1.. had the number been one greater, the number would have been completely divisible by 2 leaving no remainder.. the same is the case with 3 and 5.. so in simple words, if the number had been 1 greater, it would be a divisible by 2,3,and 5..

If the number is divisible by 2,3, and 5, then the number should be divisible by 30 .
So the number (N) is of the form-
N + 1 = 30k.. => N = 30K-1
Substitute values of k greater than 0 and you'll get three values of N.. HENCE THE ANSWER IS 3...

hope it helps!

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Re: How many positive two-digit integers have a remainder of 1 when divide   [#permalink] 23 Oct 2019, 20:29
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