@mayanksri 111111.... (27 ) times 111 × 10^24 + 111 × 10^21 + 111 × 10^18 +...... +111 × 10^3 + 111 As 1000 = 10^3 ≡ 1 mod 999. ...

@mayanksri ABCD – DBCA 1000A + 100B + 10C + D – ( 1000D + 100B + 10C + A) 999A – 999D 999( A-D) 999 is not divisible by 7 . so ( A – D) sho...

@Mayank Sum of squares of first n natural number = n ( n +1 )( 2n +1 ) /6 so (1+2²+3²+....+49²) = 49 ( 50) ( 99)/6 .

@mayanksri 12600= (2^3)* (3^2)*(5^2)*(7) powers of 2 in 50! by successive division = 50+25+12+6+3+1= 97, hence power of 2^3 will be 97/3 = 32same...

@mayanksri 71.72.73.74.……………..72020 = 7(1 + 2+ 3+ ………….2020) =72020(2020+1)/2 = 71010 x 2021 When 1010 x 2021 is divided by 4 gives remainde...

@mayanksri Please find the solution below, N = 4*8*12*....*1000 = (4*1)*(4*2)*(4*3)*.....*(4*250) = 4250*(1*2*3*...250) = 2500*250!Now 250! = 224...

@mayanksri power of 5 in 100!*200!=73. Thus number of zeros at the end of this num = 73.

@mayanksri Please find the solution below. 30! = 2^26 x 3^14 x 5^7 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1 Now the for the first non ze...

@lydiamadini Please find the solutions below. 1. The LCM of the individual intervals will give the time at which all the bells will ring togethe...

1.85 can be split into 1.000 1 +.500 1/2 +.25 1/4 +.1 1/10--------1.85 Hence 1+2+4+10=17

Hello Anu, Since 7,8 and 9 are co-prime, the number which is divisible by 7,8 and 9, must also be divisible by their L.C.M i.e. by 504. Now let's ...

Hello Anu Find the solution: 536 - 1 = 2518 - 1 = (24 + 1)18 - 1 = 2418 + 18 x 2417 + ... + 18 x 24 + 1 - 1 = 122k. OR 536 - 1 = (4 + 1)36 - 1...

@anna Let a, b, c are the 3 digits of that number. a3 = (abc + 2), b3 = (abc - 3), c3 = (abc + 3). Multiply all three equations and keep abc = x....

@anu 544 =25x17. Perfect squares divisors of 544 are: 1, 4, 16 Sum = 21

96 = 25x3; Sum of even divisors of 96 = (21+22+23+24+25)(30+31) = 248= 3600=24x32x52; Sum of odd divisors of 3600 = (30+31+32)(50+51+52) = 403 Ans...