Mathematical_Recreations-Kraitchik-2e

48 Mathematical Recreations
428 2 = 183,184, 5732 = 328,329, 7272 = 528,529, 846 2 =
715,716.
11. The corresponding problem for four-digit and eightdigit
numbers. (Hint: 101 is a prime, and 10,001 = 73 ·137,
both of which are primes.)
Answers: 912 = 8,281, 7,810 2 = 60,996,100, 9,0792 =
82,428,241, 9,9012 = 98,029,801.
12. Find a number such that each digit except zero appears
just once in the number and its square. (Thebault)
Solution: If x is the unknown number, then (x + X2) == 0
(m = 9), since the sum of the nine digits is 45. There are
two solutions of this congruence which satisfy the other requirements
of the problem: 5672 = 321,489 and 8542 =
729,316.
13. Find a number N which yields given remainders when
divided by certain given divisors.
Solution: Let p, q, r, ... be the given divisors, which we
assume to be prime to each other, and let a, b, c, ... be the
corresponding remainders.
The number M = pqr· .. is divisible
by every divisor, and each of the numbers P = qr· .. ,
Q = pr· .. , R = pqs· .. , and so on, is divisible by every divisor
but one. We must find a convenient multiple gP of P
which yields the remainder 1 on division by p. Similarly we
must find a convenient multiple hQ of Q which yields the remainder
1 on division by q, and so on. Then N = gPa + hQb
+ ... ± any multiple of M.
For example, an unknown number gives the remainders 4,
5, and 6 on division by 7, 11, and 13 respectively. Then
gP = 5(11 ·13) = 715, hQ = 4(7 ·13) -= 364, kR = 12(7 ·11) =
924, soN = 715·4+ 364:5+ 924·6 ± (n.1,D01) = 10,224 ±
n ·1,001 = 214, 1,215, and so on.
14. I count the lines of a page of my book. Counting by