Each of N, A, B and C is a base ten positive integer such that:

N = (A⁷)/7 = (B¹¹)/11 = (C¹³)/13

The minimum value of N satisfying the above set of relationships is denoted by N₀.

Determine the digital root of N₀. What is the remainder when N₀ is divided by 17?

N =A^7/7=B^11/11=C^13/13
thus
for integers t1,t2,t3 we have
A=7t1
B=11t2
C=13t3

thus we have
N = 7^6*t1^7 = 11^10*t2^11 = 13^12*t3^13
thus to minimize N we need
t1=7^a1 * 11^b1 * 13^c1
t2 =7^a2 * 11^b2 * 13^c2
t3 = 7^a3 * 11^b3 * 13^c3
this gives us

1) 7a1+6 = 11a2 = 13a3
2) 7b1 = 11b2 + 10 = 13b3
3) 7c1 = 11c2 = 13c3 + 12

solving 1 we have
11a2 = 13a3 thus a2=13k
7a1+6=143k
7a1=143k-6
a1=20k+(3k-6)/7
3k-6=7v
3k=7v+6
k=2v+2+(v/3)
v=3w
k=7w+2
a1=143w+40
a2=91w+26
a3=77w+22
thus to minimize N we take w=0 which gives
a1=40
a2=26
a3=22

solving 2+3 by similar methods we get
b1=520 b2=330 b3=280
c1=11 c2=7 c3=5

this gives us
N0 = 7^286 * 11^3640 * 13^77

now a property of the digital root is that it is equal to the numbers remainder when divided by 9 thus we need to know
7^286*11^3640*13^77 mod 9
now 7^286 mod 9 = 7
11^3640 mod 9 = 7
13^77 mod 9 = 7
thus we have
7^3 mod 9 = 1
thus the digital root of N0 is 1

for the remainder of N0 when divided by 17 we have
7^286 mod 17 = 8
11^3640 mod 17 = 16
13^77 mod 17 = 13
thus we have
8*16*13 mod 17
1664 mod 17 = 15
thus the remainder of N0 when divided by 17 is 15

Posted by Daniel on 2010-06-05 23:17:04