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 “N/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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