is 13,45,678 a perfect square
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Answers
Answer:
No, 13,45,678 is not a perfect square because it ends with the number 8.
Any number that ends with 2, 3, 7, or 8 is not a perfect square.
Hope it helps you :)
Answer:
Since 13 a + 2 b is a whole number, we can tell that the remainder we
get when we divide n^2 by 13 is the same as the remainder we get when
we divide b^2 by 13. Now we can make the following table:
b remainder when b^2 is divided by 13
0 0
1 1
2 4
3 9
4 3
5 12
6 10
7 10
8 12
9 3
10 9
11 4
12 1
This tells us a lot: a perfect square must have a remainder of
0,1,3,4,9,10, or 12 when divided by 13. That means that if a number
gives a remainder of 2,5,6,7,8, or 11 when divided by 13, then it's
not a perfect square.
Now, all of the numbers in the first column give a remainder of 2 when
divided by 13: 1991 = 153*13+2, 2004 = 1991+13 = 154*13+2, etc.
Therefore, none of the numbers in the first column can be perfect
squares!
The next column under 1992 is different. The remainder you get when
you divide 1992 by 13 is 3, so this column might contain perfect
squares. Looking at the table above, we can see that perfect squares
that give a remainder of 3 when divided by 13 are squares of numbers
that give a remainder of 3 or 9 when divided by 13.
Now, the square root of 1992 is about 44; the next bigger number that
gives 4 or 9 as its remainder is 48; 48 squared is 2304; and so 2304
is a perfect square in the second column (one way to recheck that it's
in the second column is to check that 2304-1992 is a multiple of 13).
Now continue for the rest of the columns: either eliminate them for
having a remainder other than 0,1,3,4,9,10, or 12 when divided by 13,
or find a square in the column.
This is definitely a sophisticated question. Remember to have fun with
it, and write back if you need more info. You could also look in a
library for books on number theory. Look up "modular arithmetic"
and "congruences" to find more on the method we used here.