What is the least number that when divided by 44 leaves a remainder 31

Question

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

Feb2024 · Accepted Answer

There is another way of solving such questions, the more "certain" way and it takes very much the same amount of time as answer choice elevation.
Find out the LCM of the divisors 32, 44 & 56. This value comes out as 2464 and this is the lowest number that can divide each of 32, 44 & 56 and leave no remainder. 
So, our answer is  2464 - (divisor) + (remainder) . Put in any divisor and its corresponding remainder, and you have your answer.

IanStewart · Answer

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.

Markella24 · Answer

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)

parth04red · Answer

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

stoned · Answer

Divided by 44 leaves a remainder of 31 therefore when divided by 11 number will leave a remainder of 9. only c and e satisfy this criterion.
Divided by 56 leaves a remainder of 43 therefore when divided by 7 number will leave a remainder of 1. both c and e satisfy this criterion.
Divided by 32leaves a remainder of 19 therefore when divided by 16 number will leave a remainder of 3.  only c   satisfies this criterion.

Ertucinar · Answer

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 = 	ext{lcm}(44, 56, 32) - 13 = 2464 - 13 = 2451
\boxed{2451}
\boxed{C}

kingbucky · Answer

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

What is the least number that when divided by 44 leaves a remainder 31

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What is the least number that when divided by 44 leaves a remainder 31

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