Ayushi1597
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 3Similarly, 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 4So, 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.