Integer carries Factorial

Number Theory Level 4

Find the sum of all distinct possible last digits of $\large 2^{a!} + 3^{b!}$ , where $a$ and $b$ are non-negative integers.

The answer is 25.

1 solution

Md Zuhair
Feb 15, 2017

Relevant wiki: General Diophantine Equations - Problem Solving

Here $2^{a!} + 3^{b!} = r (mod 10)$

Now For b $\leq 2$ , We can have

$2^{a!}$ can end with the numbers in the set {2,4,8,6} Now $3^{b!}$ for b $\leq 2$ We have set of values as {3,9}

Hence $2^{a!} + 3^{b!} = 5 , 7, 1 ,9 , 3 mod(10)$ ------ Set A

Again for $b > 2$ We have $3^{2} = -1 mod 10$

So For $3^6 or 3^{b!}$ for $b>0$ we have number ending with {1 , 9}

Hence with $2^{a!} + 3^{b!} = 3, 5, 9, 7 , 1 (mod 10)$ ------- Set B

Hence Our values of r = A $\cup$ B == {1,3,5,7,9}

So Sum of all r’s = {1+3+5+7+9} = $\boxed{25}$

As the number given is an odd number always that it may ends ends with $1.3.5.7.9$ .But we have to show that the expression covers all units digit as (\1,3,5,7,9).You showed that

Kushal Bose - 4 years, 3 months ago

Yes... so its correct?

Md Zuhair - 4 years, 3 months ago

Can u post it by using latex ?

Kushal Bose - 4 years, 3 months ago

@Kushal Bose – Actualy using phone now.. When I will be back on computer .. That is next day.. I will post it. But isnt it correct?

Md Zuhair - 4 years, 3 months ago

@Md Zuhair – Your approach is correct but make it clear .

Kushal Bose - 4 years, 3 months ago

@Kushal Bose – Okay.. I will try in Latex

Md Zuhair - 4 years, 3 months ago

@Md Zuhair -Actually $2^{a!}\not\equiv8\pmod{10}$ for any $a$

$2^{a!}\equiv 8\pmod{10}$ iff $a!\equiv3\pmod{4}$ which is not true for any a

for, $a=0,1$ , $2^{a!}\equiv 2\pmod{10}$

$a=2,3$ , $2^{a!}\equiv 4\pmod{10}$

$a\geq4 ,2^{a!}\equiv 6\pmod{10}$

Anirudh Sreekumar - 4 years, 2 months ago

Integer carries Factorial

The answer is 25.

1 solution

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