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jhavyom
What is the smallest number that leaves a remainder of 2 when divided by 3, 3 when divided by 4, 4 when divided by 5, 5 when divided by 6, and 6 when divided by 7?

A. 419
B. 426
C. 431
D. 433
E. 435

Let the number be n
So, n+1 is a number, which is divisible by 3, 4, 5, 6, and 7
n+1 = LCM of 3, 4, 5, 6, and 7
n+1 = 420
n = 419

Choice A is the answer.


Hi,

Could you please explain how we got to the conclusion n+1 is divisible by all the numbers (3,4,5,6,7) ?
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jhavyom
What is the smallest number that leaves a remainder of 2 when divided by 3, 3 when divided by 4, 4 when divided by 5, 5 when divided by 6, and 6 when divided by 7?

A. 419
B. 426
C. 431
D. 433
E. 435

Let the number be n
So, n+1 is a number, which is divisible by 3, 4, 5, 6, and 7
n+1 = LCM of 3, 4, 5, 6, and 7
n+1 = 420
n = 419

Choice A is the answer.


Hi,

Could you please explain how we got to the conclusion n+1 is divisible by all the numbers (3,4,5,6,7) ?

For a number (n) to be divisible by a particular divisor (d), the remainder can be any digit from 0 to d-1

Let us take a divisor.
For example, d = 3

If an integer is divided by d = 3, the remainder can be any digit from 0 to 2(i.e. d-1)

For n = 7, and d = 3, the remainder is 1
For n = 8, and d = 3, the remainder is 2
For n = 9, and d = 3, the remainder is 0 not 3
Note that, you get a number (i.e. 9) that is divisible by 3

So, for a divisor, d = 3, the remainder can be a maximum of 2 and if you add 1 more to the dividend, you will get a remainder of 0
Moreover, this new dividend will be divisible by 3

Similarly, for divisor, d = 4, the remainder can be any digit from 0 to 3

For n = 10, and d = 4, the remainder is 2
For n = 11, and d = 4, the remainder is 3
For n = 12, and d = 4, the remainder is 0 not 4

So, for a divisor, d = 4, the remainder can be a maximum of 3 and if you add 1 more to the dividend, you will get a remainder of 0
Moreover, this new dividend will be divisible by 4


So, for a number (n) that leaves a remainder of 2 when divided by 3, 3 when divided by 4, 4 when divided by 5, 5 when divided by 6, and 6 when divided by 7.
if you add 1 to that number, it will leave a remainder of 0 when divided by 3, 0 when divided by 4, 0 when divided by 5, 0 when divided by 6, and 0 when divided by 7.

Moreover, the new number obtained by adding 1 to n (i.e. n+1) will be divisible by 3, 4, 5, 6, and 7 since every digit 3, 4, 5, 6, and 7 leaves a remainder of 0 on dividing n+1.
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The only option which yields remainder 4 when divided by 5 is option A i.e 419 (A)
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