## 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 = 5^{2} + 5^{2}
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 = 5^{2} + 5^{2} + 0^{2}
803/11 = 73 = 8^{2} + 3^{2} + 0^{2}

## 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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