What is the least five-digit integer which when divided by 63, 56 and

Question

What is the least five-digit integer which when divided by 63, 56 and 42 leaves remainder 1 in each case?

(A) 10082 
(B) 10081
(C) 10071
(D) 10007 
(E) 10001

sj296 · Answer

Smallest five-digit number = 10000

Let us divide this by the L.C.M of 63, 56, 42
=> 63 = 3^2 * 7
=> 56 = 2^3 * 7
=> 42 = 2 * 3 * 7

LCM = 2^3 * 3^2 * 7 = 504

Dividing 10000 by 504 leaves a remainder 424, Subtract 424 from 10000 
=> 10000-424 = 9576

So this means 9576 is divisible by 504.

9576+504 = 10080 (which will also be divisible by 504, 63, 56, and 42)

To find the number which leaves reminder 1, we have to add 1 to 10080.

So the answer is 10081.

OA should be B

narenbs · Answer

If 1 is deducted from the correct answer, the number should be divisible by 63, 56 and 42 without leaving a remainder. Because the number (upon deduction of 1) is divisible by 63, the number should also be divisible by 9 (which is a factor of 63). Amongst the answer choices, only choice B is divisible by 9 upon deduction of 1. Hence, the answer is B.

Round2Hopeful · Answer

For a quicker method, couldnt you also notice that the divisors are multiples of 7. So, divide each of the answer choices by 7, starting from the lowest, and see which comes up with remainder 1? Seems quicker.

bv8562 · Answer

I solved it using the divisibility rules for 2, 3, 8 and 9, since all numbers are multiples of 6, 8 and 9 (42, 56, 63)

Divisibility rule for :
a) 2: Last digit must be divisible by 2
b) 3 or 9: Sum of the digits must be divisible by 3 or 9
c) 8: No. formed by the last 3 digits must be divisible by 8

When I checked for the option B i.e 10081

When divided by 2 leaves a remainder 1
Sum of digits = 10
 \frac{10}{9}  => R=1
 \frac{10}{3}  => R=1
 \frac{081}{8}  => R=1

Therefore,  our answer is B .

bumpbot · Answer

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What is the least five-digit integer which when divided by 63, 56 and

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What is the least five-digit integer which when divided by 63, 56 and

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