Question

find the sum of all two digit natural numbers which when divided by 7 yield 5 as remainder

Answer 1

The smallest 2-digit number is 10.

When divided by 7, 10 leaves quotient 1.

∴ 2 × 7 + 5 = 19 is the smallest 2-digit number which leaves remainder 5 on division by 7.

The largest 2-digit number is 99. When divided by 7, it leaves quotient 14 and remainder 1.

∴ 13 × 7 + 5 = 96 is the largest 2-digit number which leaves remainder 5 on division by 7.

Consider the numbers are in an AP.

The AP will be like 19, 26, 33,......, 82, 89, 96.

First term = T1 = 19

Common difference = d = 7

Last term = Tn = 96

No. of terms = n = (Tn - T1)/d + 1

= (96 - 19)/7 + 1

= 77 / 7 + 1

= 11 + 1 = 12

∴ There are 12 numbers.

Sum = n/2(T1 + Tn)

= 12/2 × (19 + 96)

= 6 × 115 = 690

∴ 690 is the sum.

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Answer 2

Answer:

Sum of 2-digit number which leaves 5 as remainder when divided by 7 is 690.

Step-by-step explanation:

To find: Sum of 2-digit number which leaves 5 as remainder when divided by 7.

Smallest 2-digit number = 10 and largest 2-digit number = 99

Smallest 2-digit number divisible by 7 = 14

largest 2-digit number divisible by 7 = 98

Smallest 2-digit number which leaves 5 as remainder when divided by 7 = 14 + 5 = 19

Largest 2-digit number which leaves 5 as remainder when divided by 7 = 98 + 5 - 7 = 96

From this data we get a AP : 19 , 26 , ... , 96

a = 19 , d = 7 , l = 96

$a_n=a+(n-1)d$

96 = 19 + ( n - 1 )7

7n - 7 = 77

7n = 84

n = 12

Now, Sum of all these nummber,

$S_n=\frac{n}{2}(a+l)$

$S_n=\frac{12}{2}(19+96)$

$S_n=6(115)$

$S_n=690$

Therefore, Sum of 2-digit number which leaves 5 as remainder when divided by 7 is 690.