@vishcat said:guys whats the remainder left when 5^41 is divided by 100?
25
@vishcat said:guys whats the remainder left when 5^41 is divided by 100?
vish u all the best for cat.
25
@vishcat said:guys whats the remainder left when 5^41 is divided by 100?
25...last 2 digits :)
regards
scrabbler
regards
scrabbler
@vishcat said:@TONYMBA explain pls
there is a pattern:
after 5^2%10 every 5^n%10 gives 25 as remainder...
after 5^2%10 every 5^n%10 gives 25 as remainder...
@vishcat said:guys whats the remainder left when 5^41 is divided by 100?
5powered to anything above 1 would result in a number with last two digits 25.... so...
we missed one of the previous one:
Another one: P is the vertex of a pyramid with rectangular base ABCD such that PA = 52, PB = 60 and PD = 25. Find PC.
Another one: P is the vertex of a pyramid with rectangular base ABCD such that PA = 52, PB = 60 and PD = 25. Find PC.
@vishcat said:guys whats the remainder left when 5^41 is divided by 100?
25 ?
5^41 mod 100 = 5^2*(5^3)^13 mod 100 = 25*(125)^13 mod 100 = 25*25*(25^2)^6 mod 100 = 625*(25)^6 mod 100 = 625*(25^2)^3 mod 100 = 625*625*25 mod 100 = 25*25*25 mod 100 = 25
@vishcat said:@scrabbler explain pls
Did na? Remainder when divided by 100 = last 2 digits....and last 2 digits of any power of 5 is 25...
regards
scrabbler
regards
scrabbler
@mailtoankit said:25 ?5^41 mod 100 = 5^2*(5^3)^13 mod 100 = 25*(125)^13 mod 100 = 25*25*(25^2)^6 mod 100 = 625*(25)^6 mod 100 = 625*(25^2)^3 mod 100 = 625*625*25 mod 100 = 25*25*25 mod 100 = 25
bhai 5^anything = xxxx25
where anything > 1
@vishcat said:guys whats the remainder left when 5^41 is divided by 100?
25.
@vishcat said:@Logrhythm- if 5^anything is multiplied by 50 what are the last two digits??
50
the product of ages of some teenagers is 10584000. the sum of their ages??
A>86
B>88
C>89
D>87
plz quote me..
@vishcat said:@ChirpiBird take this example as 41 x 50 x 5^41 and explain pls coz i m stuck in this step
last two digits nikalne h?
2050 * 5^41
2050*25
last two digits will be ...50 only na?
2050 * 5^41
2050*25
last two digits will be ...50 only na?
@aditi88 said:the product of ages of some teenagers is 10584000. the sum of their ages??A>86B>88C>89D>87plz quote me..
The number is 2^6 * 5^3 * 3^3 *7^2, the only way we can split it as teenagers is 15 * 15 * 15 * 14 * 14 * 16. So total is 89.
regards
scrabbler
regards
scrabbler
@aditi88 said:the product of ages of some teenagers is 10584000. the sum of their ages??A>86B>88C>89D>87plz quote me..
factorize it..
7^2*5^3*3^3*2^6
(5*3) .... 3... 15 yr olds.
(7*2).....2 .....14 yr olds
(2^4).... one 16 yr old.
rest combinations wont give u teenagers. ..
sum .. 89.
7^2*5^3*3^3*2^6
(5*3) .... 3... 15 yr olds.
(7*2).....2 .....14 yr olds
(2^4).... one 16 yr old.
rest combinations wont give u teenagers. ..
sum .. 89.
@scrabbler said:The number is 2^6 * 5^3 * 3^3 *7^2, the only way we can split it as teenagers is 15 * 15 * 15 * 14 * 14 * 16. So total is 89.regardsscrabbler
Here no of teenagers are not given ok but if total no of persons are given then can we do it by finding min value of summation of ages... when product is constant summation is min when numbers are equal..... right approach or not ???