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

1111 \times (2 + 5) + 7

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13 \times (1003 - 3) - 64 \div 8

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17 - 17 \div 17 - (17 - 1)

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It is not possibile.

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Solution

The correct option is B $13 \times (1003 - 3) - 64 \div 8$
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.

Option(a): $1111 \times (2 + 5) + 7 = 7777 + 7 = 7784$
We can see that last three digits of the number is divisible by 8 and last digit is neither 0 nor 5.
Hence the number is divisible by 8 but not by 5.

Option(b): $13 \times (1003 - 3) - 64 \div 8 = 13000 - 8 = 12992$
We can see that last three digits of the number is divisible by 8 and last digit is neither 0 nor 5.
Hence the number is divisible by 8 but not by 5.

Option(c): $17 - 17 \div 17 - (17 - 1) = 17 - 1 - 16 = 16 - 16 = 0$
We know that 0 is divisible by both 5 and 8.

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