Question

Natural number which leaves the remainder 3 on dividing by 5 and 10

Answer 1

Let the number be N.

If the number leaves remainder 3 on division by 5, then let it be 5x + 3.

If the number leaves remainder 3 on division by 10, then let it be 10y + 3.

5x + 3 = 10y + 3 = N

= 5x + 5 - 2 = 10y + 5 - 2 = N

= 5(x + 1) - 2 = 5(2y + 1) - 2 = N

5(x + 1) = 5(2y + 1) = N + 2

x + 1 = 2y + 1 = (N + 2)/5

x = 2y = ((N + 2)/5) - 1

We get an equation.

Let's give value for y.

Here, y can be the values of whole numbers.

If y = 0,

x = 2y = 2 × 0 = 0

((N + 2)/5) - 1 = 0

(N + 2)/5 = 0 + 1 = 1

N + 2 = 1 × 5 = 5

N = 5 - 2 = 3

If y = 1,

x = 2y = 2 × 1 = 2

((N + 2)/5) - 1 = 2

(N + 2)/5 = 2 + 1 = 3

N + 2 = 3 × 5 = 15

N = 15 - 2 = 13

If y = 2,

x = 2y = 2 × 2 = 4

((N + 2)/5) - 1 = 4

(N + 2)/5 = 4 + 1 = 5

N + 2 = 5 × 5 = 25

N = 25 - 2 = 23

We get an AP, i. e., 3, 13, 23, 33,......

Let's find the algebraic expression of this sequence.

First term = 3

Common difference = 13 - 3 = 10

First term - Common difference = 3 - 10 = -7

A. E. = dn + (f - d)

= 10n - 7

∴ We get the conclusion that any term which can be written in the form 10n - 7 leaves remainder 3 on division by 10 as well as 5. Or we can say that any term which leaves remainder 3 on division by 10 also leaves the same remainder 3 on division by 5.

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