IIT JEE , AIEEE Forum - Mathematics , Algebra - the last two digits of 3*3*3...... 400 times is

koms_chinda

the last two digits of 3*3*3...... 400 times is

rsashwinkumar

When 3 is multiplied two times, the last two digits is 0 9
3 times, 27
4 times, 81
5 times, 43
6 times, 29
7 times, 87
8 times, 81
9 times, 43

which repeats as 81, 43, 29, 87, 81,...
Thus when multied 4 and 8 times it is 81
Thus, when multiplied 4x25 times i.e., 100 times also it is 81.

madman

see the last digits repeat as 3,9,7,1,3,9,7,1

400 is divisible by 4

the last digit is 3

spideyunlimited

the exponent values of 3 are:
3¹ 3² 3³3⁴3⁵ 3⁶ 3⁷3⁸

= 3, 9, 27, 81, 243, 729, 2187, 6561...
The last digits occur as 3 9 7 1
so every fourth number ends with 1
so 3⁴⁰⁰ will also be after a certain number of groups of 4.
thus it ends with 1.

...workin on 2nd last digit.

ok, here is the 2nd last digit too

see every fourth number in this sequence has tens place digit has 8, 6, 4, 2, 0

3⁴ = 81

3⁸ = 6561

3¹² = 531441

3¹⁶ = 43046721

3²⁰= 3486784401

so the tens place are 8, 6, 2, 4 and 0

400 is completely divisible by 5 so it will come at the end of its group.

384, 388, 392, 396, 400

thus 3⁴⁰⁰ will have 0 as its tens place digit.

THUS, the last 2 digits of 3⁴⁰⁰ are 01.

spideyunlimited

@ashwin kumar u made a mistake 8th power of 3 is 6561 not 6581

goutam.chalasani

the last digit is 1

3^400=9^200
= (10-1)^200
expanding it using binomial theorem
each unit from ²⁰⁰C₀ 10^200 to ²⁰⁰c₁₉₈ will have their units and tens digits as zeroes

²⁰⁰C₁₉₉ 10^1 =2000
²⁰⁰C₂₀₀ 10^0 =1

(10-1)^200 =²⁰⁰C₀ 10^200 -²⁰⁰C₁ 10^199 +.....-²⁰⁰C₁₉₉ 10^1+ ²⁰⁰C₂₀₀ 10^0

except the last term each term have zeroes in units tenths and hundreths place so their addition or subtraction doesn't affect the last three places and the last term is 1 adding it to the rest of the term gives u the last three places

they are 0,0,1
hence ur tenths place is 0

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