Easy as $a,b,c,1,2,3 \space$ ?

Number Theory Level 5

Find the number of distinct ordered triples $(a,b,c)$ that satisfy the equation

$\large abc - 54 = 9a + 9b + 9c$

where $a,b,c$ are positive (not necessarily distinct) integers.

The answer is 106.

2 solutions

Brian Charlesworth
Mar 13, 2019

A quick overview of one possible solution method....

Let $x = a + 3, y = b + 3, z = c + 3$ . The given equation can then be written as

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{3}$ .

Now look for solutions with $4 \le x \le 9$ and $x \le y \le z$ . Then with, say, $x = 4$ we end up with

$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{3} - \dfrac{1}{4} = \dfrac{1}{12} \Longrightarrow 12y - 12z = yz \Longrightarrow (y - 12)(z - 12) = 144$ .

We can then look at the various factorizations of $144$ to establish values for $y,z$ and thus triples $(x,y,z)$ with $x \le y \le z$ , which are

$(4,13,156), (4,14,84), (4,15,60), (4,16,48), (4,18,36), (4,20,30), (4,21,28), (4,24,24)$ .

These translate to triples $(a,b,c)$

$(1,10,153), (1,11,81), (1,12,57), (1,13,45), (1,15,33), (1,17,27), (1,18,25), (1,21,21)$ .

To then find the number of ordered triples we then need to count all permutations of these triples. Doing this for all possible values of $x$ , we end up with a total of $\boxed{106}$ solution triples.

Note: The remainder of the ascending triples are

$(2,5,117), (2,6,42), (2,7,27), (2,9,17), (2,12,12), (3,4,39), (3,5,21), (3,6,15), (3,7,12), (3,9,9), (4,4,18), (5,5,9), (6,6,6)$ .

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Lu Min Khant - 1 year, 3 months ago

Frank Petiprin
Apr 1, 2019

def my_range4(start, end, step):
    while start <= end:
        yield start
        start += step
def my_range5(start, end, step):
    while start <= end:
        yield start
        start += step
def my_range6(start, end, step):
    while start <= end:
        yield start
        start += step
start =0
end=200
step = 1
cnt=0
for a in my_range4(start, end, step):
    for b in my_range5(start, end, step):
        for c in my_range6(start, end, step):
            left=a*b*c-54
            right =9*a+9*b+9*c
            if (abs(left-right)==0):
                cnt+=1
                print(a,b,c,left,right,'count',cnt)
print('Count Of Numbers =',cnt,'   ','Start End Step ',start,end,step)              
'''
#       Print Last 7 numbers
57 12 1 630 630 count 100
81 1 11 837 837 count 101
81 11 1 837 837 count 102
117 2 5 1116 1116 count 103
117 5 2 1116 1116 count 104
153 1 10 1476 1476 count 105
153 10 1 1476 1476 count 106

Count Of Numbers = 106    Start End Step  0 200 1

I would do it like this. The code is slightly more efficient: instead of $N^3$ loops I have just $\binom{N}{3}$ .

"""
solution of:
    https://brilliant.org/problems/easy-as-abc123-space/?ref_id=1564825
"""

import itertools

def get_weight(in_a, in_b, in_c):
    """
    returns the number of ordered triplets,
    which can be generated using the set {in_a, in_b, in_c}
    """
    tmp_len = len({in_a, in_b, in_c})
    if tmp_len == 1:
        res = 1
    elif tmp_len == 3:
        res = 6
    else:
        res = 3
    return res

N = 200
CUR_GEN = itertools.combinations_with_replacement(range(1, N), 3)
RES_SUM = sum((get_weight(a, b, c) for (a, b, c) in CUR_GEN if a*b*c-54 == 9*(a+b+c)))
print(RES_SUM)

Piotr Idzik - 1 year ago

Easy as a , b , c , 1 , 2 , 3 a,b,c,1,2,3 \space a , b , c , 1 , 2 , 3 ?

The answer is 106.

2 solutions

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Easy as $a,b,c,1,2,3 \space$ ?