Question

Can the sum of the digits of a perfect square be equal to 1970?
Please, help

Answer 1

@jim_thompson5910

Answer 2

so what mean ,,perfect square ? first of all - how many digits has ?

Answer 3

4 is a perfect square
9
144

Answer 4

ok. so this mean that x^2 = 1970 ?
but above you have wrote --- the sum of digits of a square can ....?

Answer 5

Like \(2^{1000}=(2^{500})^2\) is has 302 digits, those are 10715086071862673209484250490600018105614048117055336074437503883703510511249361‌​224931983788156958581275946729175531468251871452856923140435984577574698574803934‌​567774824230985421074605062371141877954182153046474983581941267398767559165543946‌​077062914571196477686542167660429831652624386837205668069376.
and sum of them = some number.

Answer 6

and I find another perfect square such that sum of them = 1970

Answer 7

At best there are 218 digits in this number since 1970/9 = 218.888888888889

likely there are much more digits since not all of the digits will be 9

however you look at it, this number is massive and best to use a computer to look for the number (if it even exists)

Answer 8

using 2 - get 1024
for 3 result 2187

so i think not is

Answer 9

they use log to find the number of the digits but I don't get it. Can you please explain me?

Answer 10

example: how many digits does 50,000,000 have?

we can count them out or use logs

log(50,000,000) = 7.69897000433601 which rounds to 8, so the number has 8 digits

here's how it works

log(50,000,000) = log(5*10,000,000)

log(50,000,000) = log(5) + log(10,000,000)

log(50,000,000) = log(5) + log(10^7)

log(50,000,000) = log(5) + 7*log(10)

log(50,000,000) = log(5) + 7*1

log(50,000,000) = log(5) + 7

log(50,000,000) = 7 + log(5)

log(50,000,000) = 7 + 0.69897000433601

log(50,000,000) = 7.69897000433601 (approx)

Answer 11

in general, we can factor out 10^k where k is some integer

so we can write a large number q as

q = a*10^k

where a will be some number between 0 and 10 (not including 10)

the value of log(a) will be less than 1
the value of log(10^k) will be equal to k

so

log(q) = p + k

where p is equal to log(a) and it's less than 1

Answer 12

That is the method but why does this method give us the number of digits?

Answer 13

because the logs I'm using are base 10

our counting system is base 10

each time you add on a new digit, the value of the log is increased by 1


log(1) = 0
log(10) = 1
log(100) = 2
log(1000) = 3
log(10,000) = 4
etc etc

Answer 14

oh, I got you. Thank you so much.

Answer 15

np

Answer 16

so, back to original problem, how a perfect square relate to this?

Answer 17

@xeno I think we need method than list the number out.

Answer 18

we can calculate the sum of the digits by computing the remainder when dividing by 9, ie use mod 9

here is one example using the number 1234


1234 = 1000+200+30+4
1234 = 1*1000 + 2*100 + 3*10 + 4
1234 = 1*(999+1)+ 2*(99+1) + 3*(9+1) + 4
1234 = 1*999+1+ 2*99+2 + 3*9+3 + 4
1234 = 1*999+ 2*99 + 3*9+1+2+3 + 4
1234 = (1*999+ 2*99 + 3*9)+(1+2+3 + 4)
now apply mod 9 to both sides and you'll find that
1234 = (0) + (1+2+3+4) (mod 9)
1234 = 10 (mod 9)


the terms with 999 and 99, etc will turn to 0 when we apply mod 9 since they are multiples of 9

Answer 19

now let's look at the list of integers squared mod 9

1^2 = 1 ----> 1 = 1 (mod 9)
2^2 = 4 ----> 4 = 4 (mod 9)
3^2 = 9 ----> 9 = 0 (mod 9)
4^2 = 16 ---> 16 = 7 (mod 9)
5^2 = 25 ---> 25 = 7 (mod 9)
6^2 = 36 ---> 36 = 0 (mod 9)
7^2 = 49 ---> 49 = 4 (mod 9)
8^2 = 64 ---> 64 = 1 (mod 9)
9^2 = 81 ---> 81 = 0 (mod 9)
once we get to 10, things wrap around back to 1 (since 10 = 1 in mod 9)

so when we square a number, then find the remainder mod 9, the result will either be 0, 1, 4, or 7

------------------------------------------------

now back to the original problem

we take a number x and we square it to get x^2

the question is: do the sum of the digits of x^2 equal 1970?
which is equivalent to asking "does the value of x^2, in mod 9, equal 1970?"

that's a tough question to answer, but a much easier question to answer is to compute the value of 1970 mod 9, which is equal to the sum of the digits

1+9+7+0 = 17

so 1970 = 17 (mod 9) ----> 17 = 8 (mod 9)

in the end, 1970 = 8 (mod 9) which is NOT any of the values in the list 0,1,4 or 7. So it's impossible to square a number, then add up the digits to get 1970

Answer 20

Side Note: this is a handy way to see if a number is a non-perfect square

If you have some number p, and if none of the following are true

p = 0 (mod 9)
p = 1 (mod 9)
p = 4 (mod 9)
p = 7 (mod 9)

then p is not a perfect square

Answer 21

If one of the following are true (like p = 4 mod 9), then you'll have to investigate further because p could be equal to 13, for instance

p = 13 = 4 (mod 9) but it's not a perfect square

Answer 22

Wow!! it is real tough. Thanks a ton and I need time to digest it.
I do appreciate.