Question

If the number 1 a 4 7 0 is divisible by 3, find the least and greatest value of the digit a. also justify

Answer 1

Answer:

Least value: 0 or 3 (when condition of a≠0 is given).

Greatest value: 9

Step-by-step explanation:

Sum of digits: 1 + a + 4 + 7 + 0

→12 + a

{Since the numerical value is divisible by 3, the value of a can be 0 or any digit divisible be 3 (i.e. 0, 3, 6 or 9).}

Least value of digit a: 0 (or 3, if condition of a≠0 is given.)

Greatest value of digit a: 9

Checking the Solution...

Using greatest value of digit a (9), we get,

19470

Checking the divisibility test of 3 on 19470, we get,

1 + 9 + 4 + 7 + 0

→21 (divisible by 3)

Therefore, 19470 is divisible by 3.

Hence, greatest value of digit a is 9.

Using least value of digit a (0), we get,

10470

Checking the divisibility test of 3 on 10470, we get,

1 + 0 + 4 + 7 + 0

→12 (divisible by 3)

Therefore, 10470 is divisible by 3.

Hence, least value of digit a is 0.

Using least value of digit a as 3, (when condition of a≠0 is given).

Checking the divisibility test of 3 on 13470, we get,

1 + 3 + 4 + 7 + 0

→15 (divisible by 3)

Therefore, 13470 is divisible by 3.

Hence, least value of digit a is 3, (when condition of a≠0 is given).

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