Question

Given that 37! = 13763753091226345046315979581abcdefgh0000000, determine, with a minimum of arithmetical effort, the digits a, b, c, d, e, f, g, and h. No calculators or computers allowed!

@FoolForMath up for the daily riddle /:)

Answer 1

How to solve this?

Answer 2

shhhh :-s haha

Answer 3

Nice one. Lets start with counting the number of zeroes.

Answer 4

There is 8 zeroes, and so h = 0.

Answer 5

hmmm seems about right

Answer 6

For g I know the algorithm.

Answer 7

so you can solve it /:)

Answer 8

I can solve g and h and day, I have to think about the rest.

Answer 9

Btw turns out it's one of Nick's.

Answer 10

#84.

Answer 11

hahaha yeahhh..you know him?? lol

Answer 12

No, not personally anyways.

Answer 13

a = 5, b = 8, c = 0, d = 9, e = 0, f = 2, g = 4, h = 0.

Answer 14

wow nice...howd you solve that?? hahaha

Answer 15

How do you solve that? It seems impossible to solve using paper and pen only. @mathg8

Answer 16

@mathg8 How did you solve it?

Answer 17

I know I can solve for h, g with only pencil and paper. It took a chunk of paper, but I got it.

First step is to factor \(37!\) which is surprisingly easy if you know a little trick. After that, h is easy to see, and g is also relatively easy. Beyond that, it's beyond the scope of pencil and paper unless you have lots of time.

Answer 18

How to factorize 37! ? @KingGeorge

Answer 19

How do you type inline TeX? @KingGeorge

Answer 20

Use '(' and ')' instead of '[' and ']'.

Answer 21

Thank \(\text{you!}\) @Ishaan94

Answer 22

\[\Large 37!=2^{e_1} \cdot 3^{e_2} \cdot 5^{e_3} \cdot ... \cdot 37^{e_{12}}\]Is the prime factorization. Then, to find the \(e_i\), calculate\[\Large e_i =\sum_{k=1}^\infty \lfloor{\frac{37} {p_i^k}}\rfloor\]So \[\Large e_1=\sum_{k=1}^{\infty} \lfloor {37 \over 2^k} \rfloor=34\]This is very easy to calculate since you only need to go up to the highest power of 2 such that \(2^k \leq 37\)

Answer 23

Using this, you get that \[37! =2^{34} \cdot 3^{17} \cdot 5^8 \cdot 7^5 \cdot 11^3 \cdot 13^2 \cdot 17^2 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37\]

Answer 24

Since the exponent of 5 is 8, you know that it has 8 zeroes at the end. Then, if you factor out a \((2 \cdot 5)^8\) you get rid of all trailing zeroes. What you're left with is easy to find the last digit by hand, but not any farther.