Solutions to Problem Set 5
1. Section 5.1 - Problem 2
a) The last 3 digits of 112250 are 250, which is divisible by 53 = 125 but the last 4 digits are
2250, which is not divisible by 54 = 625. Thus the largest power of 5 dividing 11250 is
53 = 125.
b) The last 4 digits of 4860625 are 0625, which is divisible by 54 = 625 but the last 5 digits are
60625, which is not divisible by 55 = 3125. Thus the largest power of 5 dividing 4860625 is
54 = 625.
c) The last 2 digits of 235555790 are 90, which is not divisible by 52 = 25 but 235555790 is
divisible by 5, thus the largest power of 5 dividing 235555790 is 51 = 5.
d) We observe that 48126953125 is divisible by 55 = 3125, as the 53125 is divisible by 3125.
Dividing 48126953125 by 55 = 3125 we get 15400625. This number is divisible by 54 =
625, but not 55 = 3125. Thus the highest power of 5 dividing 48126953125 is 55 · 54 = 59 .
2. Section 5.1 - Problem 4
A number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11.
We use this to test divisibility.
a) 1 − 0 + 7 − 6 + 3 − 7 + 3 − 2 = −1 so 10763732 is not divisible by 11.
b) 1 − 0 + 8 − 6 + 3 − 2 + 0 − 0 + 1 − 5 = 0, so 1086320015 is divisible by 11
c) 6 − 7 + 4 − 3 + 1 − 0 + 9 − 7 + 6 − 3 + 7 − 5 = 8, so 674310976375 is not divisible by 11.
d) 8 − 9 + 2 − 4 + 3 − 1 + 0 − 0 + 6 − 4 + 5 − 3 + 7 = 10, so 8924310064537 is not divisible
by 11.
3. Section 5.1 - Problem 22
We know that the total cost being x42y cents is divisible 88 and so is divisible by both 11
and 23 = 8. Thus 42y is divisible by 8 and so y = 4. As x424 is divisible by 11 we require
x − 4 + 2 − 4 = x − 6 to be divisible by 11. The only number 0 ≤ x < 10 satisfying this is
x = 6. Thus the total cost was $64.24 and so each chicken cost $64.24/88 = $0.73.
4. Section 5.5 - Problem 12
P
We use the fact that 10
i=1 ixi ≡ 0 (mod 11).
a)
0 + 2 + 3 · 9 + 4 · 8 + 5x5 + 6 · 3 + 7 · 8 + 0 + 9 · 4 + 10 · 9
≡ 5x5 + 2 + 5 − 1 + 7 + 1 + 3 + 2 (mod 11)
≡ 5x5 + 8 (mod 11) ≡ 0 (mod 11)
Thus x5 ≡ −8 · 5−1 ≡ −8 · 9 ≡ 5 (mod 11), and the missing digit is x5 = 5.
b)
9 + 2 + 15 + 20 + 20 + 12 + 7 + 16 + 9x9 + 60
≡ 9x9 + 9 + 2 + 4 + 9 + 9 + 1 + 7 + 5 + 5 (mod 11)
≡ 9x9 + 7 (mod 11) ≡ 0 (mod 11)
Thus x9 ≡ −7 · 9−1 ≡ −7 · 5 ≡ 9 (mod 11), and the missing digit is x9 = 9.
c)
x1 + 4 + 18 + 4 + 30 + 56 + 27 + 100
≡ x1 + 4 + 7 + 4 + 8 + 1 + 5 + 1 (mod 11)
≡ x1 + 8 (mod 11) ≡ 0 (mod 11)
Thus x1 = 3 is the missing digit.
5. Section 5.5 - Problem 13
P
Let S denote the sum S = 10
i=0 ixi = 21 + 8 + 40 + 54 + 72 + 45 ≡ 9 (mod 11). This
is a valid ISBN-10 with two digits transposed. Transposing the j th and k th digits changes the
sum by −jxj − kxk + jxk + kxj = (j − k)(xk − xj ). If two digits from a valid ISBN-10 were
transposed to get 0-07-289095-0, then transposing the same two digits again will give a valid
ISBN-10 again. Thus two digits in 0-07-289095-0 must satisfy (j − k)(xk − xj ) ≡ 2 (mod 11),
or equivalently (j − k)(xj − xk ) ≡ 9 (mod 11). By trial and error we find that this is satisified
by i = 8, j = 9 and in no other cases. Thus the correct ISBN-10 is 0-07-289905-0.
6. Section 6.1 - Problem 4
By Wilson’s theorem, we know that 30! ≡ −1 (mod 31). Thus we have
5!25! ≡ 25! · (−26) · (−27) · (−28) · (−29) · (−30) (mod 31)
≡ (−1)5 30! (mod 31)
≡ (−1)6 (mod 31)
≡ 1 (mod 31)
Thus 5!25! has remainder 1 when divided by 31.
7. Section 6.1 - Problem 10
By Fermat’s little theorem we have 610 ≡ 1 (mod 11). Then 62000 = (610 )200 ≡ 1200 ≡
1 (mod 11). Thus 62000 has remainder 1 when divided by 11.
8. Section 6.1 - Problem 12
By Fermat’s little theorem we have 216 ≡ −1 (mod 17). Observing that 1000000 = 106 =
2 6
26 · 56 we then have 21000000 = (216 )2 ·5 ≡ 1 (mod 17). Answer: The least postive residue of
21000000 modulo 17 is 1.
9. Section 6.1 - Problem 24
By Fermat’s little theorem we have xp ≡ x (mod p) for all x. Then
1p + 2p + 3p + · · · + (p − 1)p ≡ 1 + 2 + 3 + · · · + (p − 1) (mod p)
Since p is odd, we can rearange the terms as follows:
1+2+3+· · ·+(p−1) ≡ (1+p)+(2+(p−1))+· · ·+(
p−1 p+1
+
) ≡ 0+0+· · ·+0 ≡ 0 (mod p) .
2
2
10. Section 6.1 - Problem 26
Let rk be the least positive residue modulo 689 of 2k! for each k. These will satisfy rk+1 ≡
(mod 689). We compute rk and also (rk − 1, 689) in the hopes of finding a divisor of 689.
rkk+1
r1 = 2
(1, 689) = 1
r2 ≡ 22 ≡ 4 (mod 689)
(3, 689) = 1
3
r3 ≡ 4 ≡ 64 (mod 689)
4
2
(63, 689) = 1
2
r4 ≡ 64 = 4096 ≡ 651 = 423801 ≡ 66 (mod 689)
Thus we have found a divisor 13 of 689.
(65, 689) = 13