Challenging P6 2009 SA2 P2 MGS Q15

15. Study the number pattern below.

a) In 5⁶ what is the sum of the digits in the ones and tens place?

b) What is the sum of the last three digits in 5¹⁵?

c) What is the sum of the last four digits in 5²¹⁰?

5⁶ = 15625

5 + 2 = 7

a) In 5⁶ the sum of the digits in the ones and tens place is 7.

5⁷ = 78125

5⁸ = 390625

5⁹ = …3125

5¹⁰ = ..5625

5¹¹ = ..8125

5¹² = ..0625

…

5¹⁵ = ..8125

5 + 2 + 1 = 8

b) The sum of the last 3 digits in 5^{15 is 8.}

210 ÷ 4 = 52 r 2

5²¹⁰ = ..5625

5 + 2 + 6 + 5 = 18

c) The sum of the last 4 digits in 5²¹⁰ is 18.