Question

Find the number of two digit numbers which is divisible by 7.

Answer 1

Hint: To find the number of two digit numbers which is divisible by 7, we need to form an arithmetic progression with common difference as 7. Then we will find the ${{n}^{th}}$ term of this series which will be the highest two digit number that could be a multiple of 7. Then, we use the formula-
${{a}_{n}}$= a + (n-1) d
${{a}_{n}}$= ${{n}^{th}}$ term of this series
a = first term of the series
n = number of terms in the series
d= common difference (7 in this case)

Complete step-by-step solution:
Thus, to proceed with the above problem (to find the appropriate arithmetic series), we first find the ${{n}^{th}}$ term of this series. The ${{n}^{th}}$ term of the series would be the highest two digit number which is a multiple of 7. To find this number, we divide by 100 by 7. Doing so, we get 14 as the quotient and 2 as the remainder. Now, to get this number we subtract the remainder (that is 2) from 100 to get the ${{n}^{th}}$ term of the series. We get, 100 - 2 = 98.
The next step would be the first term of the series. This term would clearly be 14. This is because it is the smallest two digit number that is divisible by 7.
We have, the first term, a = 14 and ${{n}^{th}}$ term, ${{a}_{n}}$=98. For arithmetic progression, we have the formula-
${{a}_{n}}$= a + (n-1) d
Where, n is the required number of terms in the series and d is the common difference (which is 7 in this case)
98 = 14 + 7(n-1)
7(n-1) = 84
n-1=12
n=13
Hence, the required number of terms is 13.

Note: Another technique to arrive at the answer is to divide 100 by 7. We get 14 as the quotient. This implies that the terms are 7, 14, …, 98 (which are 14 terms). However, we have to exclude 7 from the series since it is a single digit number. Thus, the number of terms are 14-1=13.