Three-peat

Note first that finding the last two digits of an integer $N$ is equivalent to finding $N \pmod{100}$ . Also,

$\displaystyle\left(\sum_{n=1}^{33} n! \right)^{33} \pmod{100} \equiv \left(\left(\sum_{n=1}^{33} n! \right) \pmod{100}\right)^{33} \pmod{100} \equiv$

$\displaystyle\left(\sum_{n=1}^{33} (n! \pmod{100}) \right)^{33} \pmod{100}$ .

Now for $n \ge 10$ we have that $2*5*10 = 100 | n!$ , and thus $n! \pmod{100} = 0$ for $n \ge 10$ . Thus

$\displaystyle\sum_{n=1}^{33} (n! \pmod{100}) \pmod{100} \equiv \sum_{n=1}^{9} (n! \pmod{100}) \pmod{100} \equiv$

$(1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!) \pmod{100} \equiv$

$(1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80) \pmod{100} \equiv 13 \pmod{100}$ .

So we have reduced the problem to one of finding $13^{33} \pmod{100}$ .

Now as $(13,100) = 1$ we have by Euler's Totient Theorem that $13^{\phi(100)} \equiv 1 \pmod{100}$ .

Since $\phi(100) = \phi(2^{2}) \times \phi(5^{2}) = (2^{2} - 2^{1})(5^{2} - 5^{1}) = 2 \times 20 = 40$ , we have that

$13^{40} \equiv 1 \pmod{100} \Longrightarrow 13^{7}13^{33} \equiv 1 \pmod{100}$ ,

i.e., $13^{33}$ is the multiplicative inverse of $13^{7}$ modulo $100$ . Next, we conveniently find that

$13^{3} \equiv (13 \times 69) \pmod{100} \equiv 13(70 - 1) \pmod{100} \equiv$

$(910 - 13) \pmod{100} \equiv -3 \pmod{100}$ , and so

$13^{7} = 13^{3} \times 13^{3} \times 13 \equiv (-3)(-3)(13) \pmod{100} \equiv 17 \pmod{100}$ .

Next, to find the multiplicative inverse of $17$ modulo $100$ , we use the Euclidean Algorithm:

• $100 = 5 \times 17 + 15$ ,

• $17 = 1 \times 15 + 2$ ,

• $15 = 7 \times 2 + 1$ .

Then

$1 = 15 - 7 \times 2 = 15 - 7(17 - 15) = 8 \times 15 - 7 \times 17 =$

$8(100 - 5 \times 17) - 7 \times 17 = 8 \times 100 - 47 \times 17 \Longrightarrow$

$-47 \times 17 = 1 - 8 \times 100 \Longrightarrow -47 \times 17 \equiv 1 \pmod{100} \Longrightarrow 53 \times 17 \equiv 1 \pmod{100}$ .

Thus $53$ is the multiplicative inverse of $17$ modulo $100$ , and so the last two digits of the given expression are

$\Large\boxed{53}$ .

Alternatively, but much less fun, upon noting that $13^{3} \equiv -3 \pmod{100}$ we have that

$13^{33} \equiv (-3)^{11} \pmod{100} \equiv (-3)(3^{5})^{2} \pmod{100} \equiv (-3)43^{2} \pmod{100} \equiv$

$(-3)(40 + 3)^{2} \pmod{100} \equiv (-3)(1600 + 240 + 9) \pmod{100} \equiv (-3) \times 49 \pmod{100} \equiv$

$3 \times (-49) \pmod{100} \equiv 3 \times 51 \pmod{100} \equiv \boxed{53} \pmod{100}$ .

For $n > 9$ , $n!$ has $2$ and $5$ as prime factors at least twice, so $n! \equiv 0 \pmod{100}$ . As such

$\left(\displaystyle \sum_{n=1}^{33} n!\right)^{33} \equiv (1! + 2! + ... + 9! )^{33} \equiv13 ^{33} \pmod{100} \Longrightarrow$

$13^{33} = (10+3)^{33}$ = $\displaystyle \sum_{k=0}^{33} {33 \choose k} . 10^{33-k} . 3^{k}$ $\equiv 0+0+0+ ... + {33 \choose 32} . 10 . 3^{32} + {33 \choose 33} . 3^{33}$ ,

because when $k < 32$ the terms are multiples of $100$ and therefore $\equiv 0 \pmod{100}$ . This is then equal to

$33*10*3^{32} + 3^{33}$ $= 3^{33}*( 1 + 11*10 ) \equiv 3^{33}*11 \equiv 3*11*3^{32} \equiv 33*9^{16} \equiv$

$33*(10-1)^{16} \equiv 33*( {16 \choose 15}*10^{16-15}*(-1)^{15} + {16 \choose 16}) \equiv 33*(-160 + 1) \equiv$

$-33*59 \equiv -47 \equiv \boxed{53} \pmod{100}$ .

Here is a calculator free solution which uses only elementary school arithmetic.

10! and all larger factorials have two zeros as their least significant digits. (Because 2,5 and 10 all appear in the product)

So to find the last two digits of $\displaystyle\large\left(\sum_{n=1}^{33}n!\right)$ we need to take only the first 9 terms. Starting from one and multiplying sequentially by 2,3,4,5,6,7,8 and 9 and retaining only the last two digits at each step it is easy to get the final two digits of the factorials as

$01\\02\\06\\24\\20\\20\\40\\20\\80$

Adding these up shows that the last two digits of the sum of factorials are 13.

To complete the problem we need to find the last two digits of $(13)^{33}$ .

We can get most of the way by repeatedly squaring. Again we retain only the last two digits at each step.

$13^2\rightarrow 69\\13^4 \rightarrow 61 \\ 13^8 \rightarrow 21 \\ 13^{16} \rightarrow 41 \\ 13^{32}\rightarrow 81$

Then one more multiplication by 13 gives $13^{33} \rightarrow \boxed{53}$

Since any factorial from 10! onwards has at least to two trailing zeroes, we will evaluate until 9! only.

Applying modulo after evaluating and adding them up yields 13^33. 13³mod100 = 97 Tweaking a little bit, we'll tackle 97×(97²)^5 97²=9mod100 9^5=49mod100 97×49= 4753= 53mod100

Therefore, 53 is the answer.

Then there's the Ruby code way - with Ruby's BigNum feature

'
find last 2 digits of the sum n=1 to 33 of n factorial, taken to the 33rd power
'

puts 'Go'
# --- start code here ---
sum = (F[1..33]).inject(0){|s,l|s+l}
puts sum.commify

p = 1
33.times{
    p *= sum
    puts p.commify
}
# --- end code here ---
puts 'Stop'

# sum is 8,954,945,705,218,228,090,637,347,680,100,940,313
# sum ^ 33 =     26,186,507,578,944,068,851,917,969,860,217,208,810,648,149,523,729,685,779,756,163,017,095,247,959,001,264,533,989,309,171,302,889,210,362,343,362,518,712,884,488,030,263,681,175,299,268,636,558,578,687,234,022,439,639,411,059,600,226,732,471,938,464,890,757,922,186,254,070,402,734,492,603,556,116,613,946,872,500,023,224,754,984,148,665,696,278,640,979,848,178,926,311,494,870,236,560,569,567,804,289,430,833,352,829,536,770,616,908,583,120,240,873,493,538,257,716,663,624,592,171,981,968,072,663,525,917,232,646,340,186,234,544,184,664,833,734,478,235,251,198,177,103,874,023,178,122,502,647,165,927,826,672,158,381,630,152,734,353,774,739,974,006,816,698,873,844,206,346,162,162,591,850,646,364,514,488,760,635,038,337,808,404,701,935,560,147,665,482,411,102,786,007,191,420,825,088,576,286,207,306,377,078,786,955,157,512,288,476,126,300,865,376,960,497,858,118,402,006,076,367,598,670,547,064,057,229,409,367,836,311,107,603,708,762,895,013,814,730,423,093,341,199,948,549,629,089,135,048,805,554,029,158,454,574,167,538,102,427,501,873,755,061,319,786,507,430,444,039,372,019,510,150,729,298,372,465,802,487,245,057,554,419,125,345,894,487,232,565,835,212,037,626,799,216,703,026,678,063,969,286,900,511,758,442,128,804,245,971,660,730,640,454,435,305,101,201,485,424,734,181,844,299,600,584,128,271,721,359,515,827,171,213,702,287,286,873,508,491,345,799,575,255,631,608,371,310,550,770,969,823,578,479,484,776,645,129,531,694,870,960,993,475,762,463,737,807,921,695,982,099,582,220,929,409,151,029,169,199,928,718,844,610,999,122,902,726,746,183,385,426,600,662,794,773,750,292,268,070,924,549,488,091,870,567,350,553
# last 2 are 53
# correct

https://www.wolframalpha.com/input/?i=((sum+of+n!+from+1+to+33)+%5E+33+)+mod+100