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25200=2^4*3^2*5^2*7
so we have 2,4,8,16,3,9,5,25,7 as some factorsnow a divisor can be 4n+3 only for two cases
i)when it is of the form 4n+3 and has 1 prime factorii)when (4n+1)(4m+3)–>4k+3 with two prime numbers
iii) when (4n+1)(4k+1)(4m+3)=4p+3 with 3 prime factors
case i)–>3,7
2 divisorscase ii)4n+1=5,9,25
so no.of divisors representible this way=3*2-1=5 divisors
3*5
3*25
7*5
7*9
7*25case iii)4n+1—>5,9,25
we have
7*5*9 2 more cases
7*9*25
so total=2+5+2=9 divisorsanswered by Souvik Sinha Roy
7959 can be written as (80-1)59
expand it using binomal theorem 59C0800(-1)59 +59C1 801(-1)58 +…….+59C59 8059 (-1)0
So in the expansion we see that all higher terms contain 100 except for the first 2 terms
which are 59*80 and -1
= 4720-1 = 4719So last two digits is 19
answer by Harsha Vardhan
Try to solve it or let others solve . I will post sol tomorrow. :).meanwhile try other questions
abcde*4=edcba
a= 1 or 2 if >2 will give a carry over and it will result in 6 digit number
now ba should be divisible by 4 so a=2 (cant be 1)now ba can be 12,32… so b =1 or 3
also e has to end with 2 at its units place …so it can be 3 or 8 (3*4=12 and 4*8=32)if ba=12 we get e=8 therefore b=1
now d8*4 should be 12 so d can so when 4*8=32 carry over 3 now d must be 7 as 4*7=28+3 = 31
21X78*4=87X12
so X=3,5,6,9
so trial and error methods se X=9so answer is 21978*4=87912
Sum of digits = 27Answer given by Vaibhav AutadeLet the amount with amar be X.
so Others(Bhavan , chetan and dinesh) have amount = 4X
Total amount = X + 4X = 5X = 150 –> X= 30
Hence Amar has Rs. 30
start removing smallest digits 1,2,3,5 ..
you will get the largest possible number as 9876
hence largest omitted digit is 5.
Let the cheque amount be x rupees and y paise.
so total amount of cheque is 100x +y paise.
and transposed amount = 100y + x paisa.
so 100y+x – 50 = 3(100x+y)
97y = 50 + 299x
Also since y is paisa so, 0<y<100
–> 0<=50+299x<=99×97 –> 47<= 299x <=96673
Preparation for 3 months with cool mind and proper planning is required and one can surely fetch score above 700
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