## Number system

Hello all, I got 2 problems on Numbers. Are there any best possible solutions for these 2 problems? 1) A 3 digit number N is such that it is divisible by 11. Also, N/11 is equal to the sum of squares of three digits of N. N can take ......... value(s) 2) Let N=1000!. N is divided successively by distinct numbers of the form X^x. Where x is a prime number. What is the maximum number of divisions that can be performed, before getting a non-zero remainder? Thanks in advance.

## Nice Question

The number can be of the form aa0 or x0y where x + y = 11 Aa0 / 11 = a0 = sum of the squares of a, a and 0(given condition) Possible values 110/11 = 10 220/11=20 330/11=30 440/11=40 550/11 = 50 = 52 + 52 660/11=60 770/11=70 880/11=80 990/11=90 For the other form x0y 209/ 11 = 19 308/11 = 28 407/11 = 37 506/11 = 46 605/11 = 55 704/11 = 64 803/11 = 73 = 64 + 9 902/11 = 82 So there are two possible numbers 803 and 550 If the number is of the form xyz then x + y + z should be equal to 11 or 22 (any multiple of 22 but anything more than 22 (i.e 33 ) is not possible for 3 digit numbers) or x-y+z=0 => x+z =y But none of the number with above form will satisfy the condition â€œN/11 is equal to the sum of squares of three digits of Nâ€

## but the condition that N/11=

but the condition that N/11= sum of the squares of all the 3 digits of that no. is not satisfied.........consider for ex;- 110/11=10, which is not equal to(1*1+1*1+0*).........so what do you say????????????

## Thats what i had tried to

exactly what i had tried to show with examples There only 2 possible numbers 550/11 = 50 = 52 + 52 + 02 803/11 = 73 = 82 + 32 + 02

## Q)What is the remaider of

Q)What is the remaider of (4*4!+5*5!+.....20*20!/64)..try n tell me

## anything more than 8 x 8! is

anything more than 8 x 8! is divisible by 64 So it s like finding the remainder of (4 x 4! + 5 x 5! + 6 x 6! + 7 x 7!) for 4 x 4! reminder is 32 for 5 x 5! reminder is 24 for 6 x 6! reminder is 32 for 7 x 7! reminder is 16 So the answer is 40.

## But dont you think finding

But dont you think finding all the 4 reminders is time taking is there any other simple and easy way

## There might be some easier

There might be some easier method but its not that time taking. let me tell you how i found all the reminders for 4 x 4! = 96 so reminder 32 for 5 x 5! i know 5! is 120 so when divided by 64 the negative reminder is -8 (reminder 56) so I have to find the reminder for -40 and thats 24... Is it that tough Similarly for 6 x 6! you should remember the value of 6! else follow as above i. e 6 x 6 ! = 36 x 5! ; reminder for 5! is 24 so we have to find the reminder for 36 x 24 same as the reminder of 72 x 12 same as reminder of 8 x 12 and the reminder of 96 is 32 i don't think all of the above wld take more than 1 min and that w/o pen and paper

## shortcut to number problem

4*4! + 5*5! +.....+20*20!= (5-1)*4! + (6-1)*5! +.... +(21-1)*20! =(5!-4!) + (6!-5!) +......(21!-20!) =21!-4! now 21!=21*20.....*8*...*4*...*2*1 is divisible by 64 as8*4*2is 64 Now to the second part -4! as -4!=-24 here 64-24=40 remainder as -24 means 24 less than 64;

## what is the remeinder of

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