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Positive integer A > 10,000 and B > 20,000. When A is divided by 237

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Bunuel
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Positive integer A > 10,000 and B > 20,000. When A is divided by 237 [#permalink] New post   10 Feb 2022, 05:45
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Positive integer A > 10,000 and B > 20,000. When A is divided by 237, the remainder is 85. When B is divided by 237, the remainder is 41. What is the remainder when (2A + 3B) is divided by 237?

A. 0
B. 56
C. 123
D. 170
E. 211
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BrentGMATPrepNow
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Re: Positive integer A > 10,000 and B > 20,000. When A is divided by 237 [#permalink] New post   10 Feb 2022, 07:57
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Bunuel wrote:
Positive integer A > 10,000 and B > 20,000. When A is divided by 237, the remainder is 85. When B is divided by 237, the remainder is 41. What is the remainder when (2A + 3B) is divided by 237?

A. 0
B. 56
C. 123
D. 170
E. 211


Key property: We can take ANY possible values of A and B (that satisfy the given information) and use them to determine the remainder when (2A + 3B) is divided by 237.
So we can ignore the information that tells us A > 10,000 and B > 20,000, and just use the smallest possible values of A and B.

Another property: 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.

Given: When A is divided by 237, the remainder is 85
So, 85 is the smallest possible value of A.

Given: When B is divided by 237, the remainder is 41
So, 41 is the smallest possible value of B.

Question: What is the remainder when (2A + 3B) is divided by 237?
2A + 3B = 2(85) + 3(41) = 170 + 123 = 293
293 divided by 237 equals 1 with remainder 56

Answer: B

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Re: Positive integer A > 10,000 and B > 20,000. When A is divided by 237 [#permalink] New post   11 Feb 2022, 10:26
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From the question stem we can deduce:

A = 237Q + 85
B = 237K + 41

where K and Q are some numbers.

The question asks remainder of (2A + 3B) divided by 237:

2A + 3B = 2(237Q + 85) + 3(237K + 41)

We know that 237Q and 237K will give us remainder 0 when divided by 237 as they will be multiples of 237.

Hence this leaves us with what will the remainder be for:

((2*85) + *(3*41) ) /237

= (170 + 123) / 237

= 293/237

= 1 Remainder 56

==> B
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Re: Positive integer A > 10,000 and B > 20,000. When A is divided by 237 [#permalink] New post   08 Sep 2022, 21:10
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When A is divided by 237, the remainder is 85

Theory: Dividend = Divisor*Quotient + Remainder

A -> Dividend
237 -> Divisor
a -> Quotient (Assume)
85 -> Remainders
=> A = 237*a + 85 = 237a + 85

When B is divided by 237, the remainder is 41

Let quotient be b
=> B = 237*b + 41 = 237b + 41

What is the remainder when (2A + 3B) is divided by 237

2A + 3B = 2*(237a + 85) + 3*(237b + 41) = 237*2a + 170 + 237*3b + 123 = 237*(2a + 3b) + 237 + 56 = 237*(2a + 3b + 1) + 56

=> 2A + 3B when divided by 237 gives 2a + 3b + 1 as quotient and 56 as remainder.

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Remainders

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Positive integer A > 10,000 and B > 20,000. When A is divided by 237 [#permalink] New post   08 Sep 2022, 21:48
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Bunuel wrote:
Positive integer A > 10,000 and B > 20,000. When A is divided by 237, the remainder is 85. When B is divided by 237, the remainder is 41. What is the remainder when (2A + 3B) is divided by 237?

A. 0
B. 56
C. 123
D. 170
E. 211


Recall what division is - grouping. When we divide A marbles by 237, we make groups of 237 and we have 85 leftover. If we have 2A marbles, we will have 2 groups of 85 i.e. 170 marbles leftover.
When we divide B marbles by 237, we make groups of 237 and we have 41 marbles leftover. If we have 3B, we will have 3 groups of 41 marbles leftover i.e. 123 marbles.
So with 170 + 123 = 293 marbles leftover, we can make another group of 237 marbles. So eventually we have only 293 - 237 = 56 marbles leftover
Answer (B)

Check the details of this concept here: https://youtu.be/A5abKfUBFSc
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Positive integer A > 10,000 and B > 20,000. When A is divided by 237 [#permalink]
08 Sep 2022, 21:48
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