Question

Find the number(s) that is/are divisible by $8$ but not by $5.$

• A. $1111\times (2+7)+9$
• B. $17\times (17-3)-48÷8$
• C. $8\times 50-({14}^2-6)$
• D. None of the above.

Answer 1

Answer: (A), (B)

$17\times (17-3)-48÷8$

Divisibility by 8: A number whose last three digits are divisible by 8, is also divisible by 8.

Divisibility by 5: A number whose last digit is '0' or '5', is divisible by 5.

$\underset{–}{\text{Option(a)}}:$ $1111\times (2+7)+9=9999+9=10008$

We can see that the number's last three digits are divisible by 8, and the last digit is neither 0 nor 5.

Hence, the number is divisible by 8 but not by 5.

$\underset{–}{\text{Option(b)}}:$ $17\times (17-3)-48÷8=238-6=232$

We can see that the number is divisible by 8, but the last digit is neither 0 nor 5.

Hence, the number is divisible by 8 but not by 5.

$\underset{–}{\text{Option(c)}}:$ $8\times 50-({14}^2-6)=400-(196-6)=400-190=210$

$210$ is divisible by 5 but not divisible by 8.
Hence, it is not a correct option.