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Exponential Difficulties 3 (Posted on 2010-06-05) Difficulty: 3 of 5
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?

No Solution Yet Submitted by K Sengupta    
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analytical solution Comment 1 of 1

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
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