Question

WHat are last two digits of product 5×5×5×... 5 where factor 5 appears 100 times?​

Answer 1

Answer:

5

Step-by-step explanation:

Because if we multiply 5 by 5 then we get 25 as answer whose last digit is 5 and again if we multiply 25 by 5 we get 125 whose last digit is again 5 . In this way, how many times we multiply the number 5 by itself we get 5 as last digit.

Answer 2

Solution :-

Concept :- for a number with unit digit as 5 ,

• If power is even , last two digits will be 25 {5^2n = 25 in last , 45^2n = 25 in last , 195^2n = 25 in last .}
• If power is odd , but ten's digit number is even , last two digits will be 25 . {25³ = 15625, 45³ = 91125 , 65⁹ = 25 in last, 85^17 = 25 in last .
• If power is odd , and ten's digit number is also odd , last two digits will be 75 . { 15³ = 75 in last, 35³ = 75 in last, 55⁹ = 75 in last . }

So,

→ 5 * 5 * 5 * 5 ___________ 100 times

→ 5¹⁰⁰

since 100 ÷ 2 = remainder 0 . Power is an even number .

then,

→ 5^(2n)

→ 25 in last .

therefore, the two digits of given product will be 25 .

Learn more :-

what is the remainder when 61^64846 is divided by 63?

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