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gmatophobia
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There are more systematic ways to do questions like this, but we don't need to bother with those, because we know one of the answer choices is correct. We get an odd remainder dividing by 32, so our number must be odd, leaving B, C or E as the answer. We can test 603: since 640 - 64 = 576 is a multiple of 32, the remainder when we divide 603 by 32 is not 19 (it is 603 - 576 = 27), so answer E is wrong. The remaining two answers are 26 apart, so they'll definitely give different remainders when we divide by any of the three numbers in the question (since those numbers are all larger than 26), so we now can just divide one of the two remaining answer choices by one of 32, 44 or 56, and find the remainder, and if the remainder is right, the answer is right. Or we can locate a multiple of, say, 44, that's nearby: since 44*50 = 2200, then 44*55 = 2200 + 220 = 2420, and when we divide 2451 by 44, we see we get a remainder of 31, so 2451 must be the answer.
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Looking back at my suggested solution for the exercise it was not correct, so I had to remove it. Sorry for the confusion.
(for those who read it... something that is divisible by a smaller factor of the numbers given e.g. divisible by 8 is not necessarily also divisible by 32)
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Markella24
We notice here that all 44, 56 and 32 are multiples of 4. So, to help our calculations we can rephrase as follows:
What is the least number that when divided by 11 leaves a remainder 9, when divided by 14 leaves a remainder 11, and when divided by 8 leaves a remainder 3? (notice here that I calculated new remainders based on the new divisors so e.g. 11 divides 31 and leaves a remainder of 9).

Next, I am attacking the answer choices starting from choice C (middle value). In this case, it does not really matter where you start but it is usually helpful to start in the middle.

I choose to test with 11.
So, I take my answer choice and substract the remainder i.e. 2451-9=2442
According to the divisibility rules if a number is a multiple of 11 then by substracting the value of the last digit from the rest of the number you should get a multiple of 11 and this is a repeatable process.

Therefore,
244-2=242, which is a multiple of 11! However, if you were not able to identify that, you could repeat the process and get:
24-2=22, here you should easily identify that this is a multiple of 11. And there you have your answer C.

Shouldn't the remainder be 1 when the number is divided by14? 43 = 14*3 + 1.
or am i missing something here?

and good take on divisibility by 11. This method seems simpler than the difference of odd and even values rule, provided i can remember it :)
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\(\\
N \equiv 31 \pmod{44}, \quad N \equiv 43 \pmod{56}, \quad N \equiv 19 \pmod{32}\\
\)
By adding 13 to both sides, we transform the system into:
\(\\
N + 13 \equiv 0 \pmod{44}, \quad N + 13 \equiv 0 \pmod{56}, \quad N + 13 \equiv 0 \pmod{32}\\
\\
\)
Since the least common multiple of 44, 56, and 32 is 2464, we get:
\( \\
N = \text{lcm}(44, 56, 32) - 13 = 2464 - 13 = 2451\\
\boxed{2451}\\
\boxed{C}\\
\)
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What is the least number that when divided by 44 leaves a remainder 31, when divided by 56 leaves a remainder 43, and when divided by 32 leaves a remainder 19?

Lets call the number N,
N = 44k' +31 = 44k - 13
N = 56m - 13
N = 32j - 13

N + 13 = multiple of 44, 56 and 32 = multiple of LCM(44,56,32) = 11*32*7 = multiple of 2464
N = 2464b -13 (b = 1 gives the least value for N)

i.e. gives N = 2451
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gmatophobia
What is the least number that when divided by 44 leaves a remainder 31, when divided by 56 leaves a remainder 43, and when divided by 32 leaves a remainder 19?

A. 2464
B. 2477
C. 2451
D. 616
E. 603



The question says:

A number leaves remainder 31 when divided by 44.

This means:

N = 44a + 31

where a is some whole number.

Similarly,

A number leaves remainder 43 when divided by 56.

So:

N = 56b + 43

And,

A number leaves remainder 19 when divided by 32.

So:

N = 32c + 19



Now notice something interesting:

44a+31+13=44a+44=44(a+1)

which is a multiple of 44.

Similarly,

56b+43+13=56b+56=56(b+1)

which is a multiple of 56.

And,

32c+19+13=32c+32=32(c+1)

which is a multiple of 32.



So if we define

M=N+13

then M is simultaneously:

* a multiple of 44
* a multiple of 56
* a multiple of 32

Therefore M must be a common multiple of all three numbers.

The smallest such number is their LCM.

So the smallest possible value of M is

M=2464

But

M=N+13

Therefore

N=2464-13

N=2451
Answer ( C )
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The shortcut is to notice that each remainder is 13 less than its divisor. So if you add 13 to the number, it becomes divisible by 44, 56, and 32. Then just find the LCM:
LCM(44, 56, 32) = 2464
So the number is 2464 − 13 = 2451.
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