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Reminder Problem

Expert replies
Source: — Problem Solving |

by GMATGuruNY » Sun Sep 25, 2016 3:24 pm
Mo2men wrote:Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16 is 4?

a) 137
b) 1361
c) 140
d) 157
e) 172
7
12 = 2*2*3.
16 = 2*2*2*2.

As the factors in blue illustrate:
To be divisible by 7, 12, and 16, an integer must include at least one 7, four two's and one 3.
Thus, the LCM of 7, 12 and 16 = (2�)(3)(7) = 336.

Implication:
To leave a remainder of 4 when divided by 7, 12 and 16, an integer must be 4 MORE THAN A MULTIPLE OF 336.
Thus, the integer must be in the following form:
336a + 4, where a is a nonnegative integer.
If a=5, we get:
(336*5) + 4 = 1684.

The integer in red is the greatest possible integer less than 1856 that will leave a remainder of 4 when divided by 7, 12 and 16.
1856-1684 = 172.
Thus, to yield an integer that will leave a remainder of 4 when divided by 7, 12 and 16, the smallest value that must be subtracted from 1856 is 172.

The correct answer is E.
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by GMATGuruNY » Mon Sep 26, 2016 8:13 am
Mo2men wrote:Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16 is 4?

a) 137
b) 1361
c) 140
d) 157
e) 172
An alternate approach is to PLUG IN THE ANSWERS.

A value that yields a remainder of 4 when divided by 16 must be of the following form:
16a + 4 = (multiple of 16) + 4 = (multiple of 4) + 4 = multiple of 4.
Thus, when the correct answer choice is subtracted from 1856, the result must be a multiple of 4.
Since 1856 is EVEN, subtracting A, B, or D will yield EVEN - ODD = ODD.
Since the result will be ODD -- and thus not a multiple of 4 -- eliminate A, B and D.

The smaller of the two remaining answer choices is C.

Answer choice C:
1856 - 140 = 1716.
1716/7 = 245 R1.
Since dividing by 7 does not yield a remainder of 4, eliminate C.

The correct answer is E.
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by Mo2men » Mon Sep 26, 2016 12:55 pm
GMATGuruNY wrote:
Mo2men wrote:Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16 is 4?

a) 137
b) 1361
c) 140
d) 157
e) 172
An alternate approach is to PLUG IN THE ANSWERS.

A value that yields a remainder of 4 when divided by 16 must be of the following form:
16a + 4 = (multiple of 16) + 4 = (multiple of 4) + 4 = multiple of 4.
Thus, when the correct answer choice is subtracted from 1856, the result must be a multiple of 4.
Since 1856 is EVEN, subtracting A, B, or D will yield EVEN - ODD = ODD.
Since the result will be ODD -- and thus not a multiple of 4 -- eliminate A, B and D.

The smaller of the two remaining answer choices is C.

Answer choice C:
1856 - 140 = 1716.
1716/7 = 245 R1.
Since dividing by 7 does not yield a remainder of 4, eliminate C.

The correct answer is E.
Hi Mitch,

Thanks for both answers. In the second solution, I think we can examine 1716 more quickly if we figure out that there is no reminder when 1716 is divided by 12 (R=0) as 1716 divides 12 evenly. So Eliminate C

Am I right?
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by GMATGuruNY » Tue Sep 27, 2016 2:25 am
Mo2men wrote:In the second solution, I think we can examine 1716 more quickly if we figure out that there is no reminder when 1716 is divided by 12 (R=0) as 1716 divides 12 evenly. So Eliminate C

Am I right?
Yes.
Another way to eliminate C is as follows:
Since the sum of the digits of 1716 is a multiple of 3 -- 1+7+1+6=15 -- 1716 is a multiple of 3.
Since the last two digits of 1716 form a multiple of 4 -- 16 -- 1716 is a multiple of 4.
Since 1716 is divisible by both 3 and 4, 1716 is a multiple of 12.
Thus, dividing 1716 by 12 will not leave a remainder of 4.
Eliminate C.
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I unlock the best way for YOU to solve problems.

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by Matt@VeritasPrep » Thu Sep 29, 2016 7:06 pm
We don't need much text to explain this, it can be done with a few equations.

We can rephrase the prompt as

1856 - x - 4 = (some common multiple of 7, 12, and 16)

The LCM of 7, 12, and 16 is 336, so

1856 - 4 - x = 336*(some integer)

1852 - x = 336*some integer

336 * 6 will be way > 1800, so we want 336 * 5.

1852 - x = 336*5

x = 172
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by Matt@VeritasPrep » Thu Sep 29, 2016 7:08 pm
Addendum: we want the prompt to specify that the number is POSITIVE. (I assumed that, given the answer choices, but without that clarification we'd be able to use arbitrarily small negative numbers.)
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