Powers and Factorials

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Find the number of ordered triples of nonnegative integers $(a,b,k)$ such that $1+2^a+4^b=k!$

The answer is 2.

2 solutions

Daniel Chiu
Dec 14, 2013

Note that if $a\ge 2$ or $b\ge 1$ , then the left side is not divisible by 4 and $k\le 3$ . Therefore, if $k>3$ , then $a<2$ and $b<1$ , and the maximum value of the LHS is $1+2+1=4$ , a contradiction, so $k\le 3$ .

If $b=0$ , then $2+2^a=k!$ Since $k\le 3$ , we can easily see the only solution here is $(2,0,3)$ .

If $b=1$ , then $5+2^a=k!$ Since $k\le 3$ , we can easily see the only solution here is $(0,1,3)$ .

If $b>1$ , then the LHS is at least 18, but since $k\le 3$ , this is not possible.

Therefore, there are $\boxed{2}$ solutions.

Nice solution. I did it this way, pretty similar though.

If $(a,b)>0$ then left side is odd, but the only odd factorial number is $1$ , observing the expression, we see that it's not possible. Hence exactly one of $a$ and $b$ must be $0$ .

Either way we have an expression like $2^m+2=k!$ . If $m=0,1$ , we obtain $3$ and $4$ which are not factorial numbers, if $m>1$ , then $2^m+2\equiv 2 \pmod{4}$ . So we need to check the factorial numbers which don't have $4$ as a factor. And so $k\le 3$ .

Fixing $a=0$ , we have $2^{2b}+2=k!$ . Checking $3$ cases, we obtain $(a,b,k)=(0,1,3)$ . Fixing $b=0$ we have $2^a+2=k!$ . Again checking $3$ cases, we obtain $(a,b,k)=(2,0,3)$ .

Hence, there are exactly $\fbox{2}$ solutions.

At first glance, I immediately tried $0$ since the left side was odd and there's only one odd factorial which doesn't work. Then I noticed the word non-negative . :3

Jubayer Nirjhor - 7 years, 6 months ago

John Ashley Capellan
Dec 14, 2013

To simplify the discussion:

Let us take some factorials (I'm not choosing in which numbers I'm going to manipulate).

Suppose k is greater than or equal to 5. This means that... taking their modulo 10's gives 0. But in the left-hand side of the expression, taking their modulo 10's are not all zeroes in all cases.

Counter-example: 1. Suppose a = 0 (for the expression to be even) and b be any number except zero. Taking modulo 10 will give either 6 or 8. This is by powers of 4. This does not satisfy.

Suppose a be any number except zero but b is zero. taking modulo 10 will give 0 if and only if a is congruent to 3 modulo 4. We can show that for any a congruent to 3 cannot be expressed as a factorial simply by computing modulo 3 (Remembering the definition of a factorial). For any a congruent to 3 (modulo 4), the expression is always congruent to 1 (modulo 3) which implies that it cannot be divisible by 3 which violates basically the condition of being expressed as a factorial. Some numbers are divides this number but a basic counterexample of 3 finishes this proof.

We are now finished making proof for k greater than or equal to 5. Now let's use case-to-case basis for k between 0 and 4.

For k = 0, 1. There is not integral solutions for 2^a + 4^b = 0.

For k = 2. When 2^a + 4^b = 1, there are rational solutions but in the given, negative integers are not accepted.

For k = 3. When 2^a + 4^b = 5, trivially, there are two solutions: (a, b, k): (2, 0, 3) and (0, 1, 3)

For k = 4, When 2^a + 4^b = 23, one of the numbers must be odd in order for the number maybe a power of 2. Plugging a or b = 0, still there are no integral solutions since 22 is not an integral power of 2.

Hence, there are two ordered pairs of non-negative integers.

Powers and Factorials

The answer is 2.

2 solutions

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