Question

How many three digit numbes are such that when divided by 7 leave a remainder 3 in each casel? ​

Answer 1

$\sf{\underline{Solution:-}}$

• first three digit number is 100

• if 100 divided by 7 then remainder is 2

so,

101 is the first number who when divides by 7

→ remaining 3

→ next number must be greater by 7

→ so it is 108

→ so series is

★ 101, 108, 115...

• last three digit number is 999 when divides by 7 remaining 5

• so 997 is the number when divides by 7

remaining 3

→ so series is

→ 101, 108, 115 ..... 997

† here a = 101

• d = 7
• ᵗn = 997
• n = ?

★ Formula = ᵗn = a + ( n - 1) d

→997 = 101 + (n-1)7

→ 997 - 101 = (n- 1) x 7 !!

→ (n-1)* 7 = 896

→ n-1 = 896/7

→ n - 1= 128

→ n=128 + 1

→ n = 129

$\sf{\underline{Hence:-}}$

• The required number is 129 which is divided by 7 and leaves remainder 3

Answer 2

Question :–

How many three digit numbes are such that when divided by 7 leave a remainder 3 in each case ?

Solution :–

If we divide 100 by 7 so, we get 2 as a remainder.

That's why, we divided 101 by 7 then we get 3 as a remainder.

So,

• 101, 108, 115....

For the last term,

If we have 999 by 7 then we didn't get 3 as a remainder. So,

• 999 - 3 = 997

If we subtract 3 in the last 999 then we get 997.

And if we divide 997 by 7 then the remaining number is 3.

Now, the series will be :–

• 101, 108, 115..............997

These are in A.P.

• First term (a₁) = 101
• Second term (a₂) = 108
• Third term (a₃) = 115
• Last term $(a_{n})$ = 997

Common difference (d) = a₂ - a₁ = a₃ - a₂

→ 108 - 101 = 115 - 108

→ 7 = 7

• Common difference (d) = 7

→ $\sf a_{n}$ = a₁ + (n - 1)d

→ 997 = 101 + (n - 1)7

→ 997 - 101 = (n - 1)7

→ 896 = (n - 1)7

→ $\sf \frac{896}{7}$ = n - 1

→ 128 = n - 1

→ 128 + 1 = n

→ 129 = n

Hence,

129 is the number which is divided by 7 and leave remainder 3.