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# The Easiest Complicated Problem Ever!!

If $$A_{n}=n^{(n+1)^{(n+2)^{(n+3)^{...^{2012^{2013}}}}}}$$, what is the last digit of $$A_{1}+A_{2}+A_{3}+...+A_{2013}$$?

Note by Ignatius Alvin Wijaya
4 years, 2 months ago

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## Comments

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Solution:

If n is congruent to 5 modulo 10, then the last digit is always 5. If n is congruent to 0 modulo 10, then the last digit is always 0. If n is congruent to 6 modulo 10, then the last digit is always 6. If n is congruent to 1 modulo 10, then the last digit is always 1. If n is congruent to 9 modulo 10, then the last digit is always 1 since 9 is congruent to -1 modulo 10 and exponent raised in the given is always even. If n is congruent to 2 modulo 10, Case 1: If the succeeding exponent is congruent to 3 modulo 4, then the last digit is 2. Case 2: If the succeeding exponent is congruent to 1 modulo 4, then the last digit is 2. If n is congruent to 3 modulo 10, Case 1: If the succeeding exponent is congruent to 0 modulo 4, then the last digit is 1. Case 2: If the succeeding exponent is congruent to 2 modulo 4, then the last digit is 1. If n is congruent to 4 modulo 10, Case 1: If the succeeding exponent is congruent to 3 modulo 4, then the last digit is 4. Case 2: If the succeeding exponent is congruent to 1 modulo 4, then the last digit is 4. If n is congruent to 7 modulo 10, Case 1: If the succeeding exponent is congruent to 0 modulo 4, then the last digit is 1. Case 2: If the succeeding exponent is congruent to 2 modulo 4, then the last digit is 1. If n is congruent to 8 modulo 10, Case 1: If the succeeding exponent is congruent to 3 modulo 4, then the last digit is 8. Case 2: If the succeeding exponent is congruent to 1 modulo 4, then the last digit is 8.

So adding the (remainders) gives 201(0 + 1 + 8 + 1 + 6 + 5 + 4 + 1 + 2 + 1) + 6 = 201(29) + 6 is congruent to 9 + 6 is congruent to 5 modulo 10. Hence the last digit is 5.

- 4 years, 2 months ago

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Proof why when n is raised continuously say n^(n+1)^(n+2)....2013 gives the same last digit. As mentioned earlier: If n is congruent to 5 modulo 10, then the last digit is always 5. If n is congruent to 0 modulo 10, then the last digit is always 0. If n is congruent to 6 modulo 10, then the last digit is always 6. If n is congruent to 1 modulo 10, then the last digit is always 1.

Acronym used: Exponent is congruent to - EICT How about the numbers ending in 9: Possible last digits when raised exponentially: 9 (if EICT 1 modulo 2) and 1(if EICT 0 modulo 2) Say the succeeding exponent is divisible by 2 (i.e. 19^20^21^...^2013). Since 20 is congruent to 0 modulo 2, then the last digit is 1. This is always the case because for every number ending in 9 gives succeeding exponents which has even base.

Numbers ending in 8: Possible last digits when raised exponentially: 8 (if EICT 1 modulo 4), 4 (if EICT 2 modulo 4), 2 (if EICT 3 modulo 4), and 6 (if EICT 0 modulo 4) Say the succeeding exponent is congruent to 1 modulo 4 (i.e. 8^9^10^11^...^2013). Since 9 is congruent to 1 modulo 4, then the last digit is 8. Assume that the exponent is congruent to 3 or -1 modulo 4 (i.e. 18^19^20^...^2013). Since 19 is congruent to -1 modulo 4, then the last digit is also 8.

Numbers ending in 7: Possible last digits when raised exponentially: 7 (if EICT 1 modulo 4), 9 (if EICT 2 modulo 4), 3 (if EICT 3 modulo 4), and 1 (if EICT 0 modulo 4) Say the succeeding exponent is congruent to 0 modulo 4 (i.e. 7^8^9^10^...^2013). Since 8 is congruent to 0 modulo 4, then the last digit is 1. Assume that the exponent is congruent to 2 modulo 4 (i.e. 17^18^19^...^2013). Since 18 is congruent to 2 modulo 4 which implies that 18^19^20^...2013 is congruent to 2^19^20^...^2013 is congruent to 0 modulo 4, then the last digit is also 1.

Numbers ending in 4: Possible last digits when raised exponentially: 4 (if EICT 1 modulo 2) and 6(if EICT 0 modulo 2) Say the succeeding exponent is congruent to 1 modulo 2 (i.e. 4^5^6^...^2013). Since 5 is congruent to 1 modulo 2, then the last digit is 4. This is always the case because for every number ending in 4 gives succeeding exponents which has base congruent to 1 modulo 2.

Numbers ending in 3: Possible last digits when raised exponentially: 3 (if EICT 1 modulo 4), 9 (if EICT 2 modulo 4), 7 (if EICT 3 modulo 4), and 1 (if EICT 0 modulo 4) Say the succeeding exponent is congruent to 0 modulo 4 (i.e. 3^4^5^6^...^2013). Since 4 is congruent to 0 modulo 4, then the last digit is 1. Assume that the exponent is congruent to 2 modulo 4 (i.e. 13^14^15^...^2013). Since 14 is congruent to 2 modulo 4 which implies that 14^15^16^...2013 is congruent to 2^15^16^...^2013 is congruent to 0 modulo 4, then the last digit is also 1.

Numbers ending in 2: Possible last digits when raised exponentially: 2 (if EICT 1 modulo 4), 4 (if EICT 2 modulo 4), 8 (if EICT 3 modulo 4), and 6 (if EICT 0 modulo 4) Say the succeeding exponent is congruent to 3 or -1 modulo 4 (i.e. 2^3^4^5^...^2013). Since 3 is congruent to 0 is congruent to -1 modulo 4, then the last digit is 2. Assume that the exponent is congruent to 1 modulo 4 (i.e. 12^13^14^...^2013). Since 13 is congruent to 1 modulo 4, then the last digit is also 2.

- 4 years, 2 months ago

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I don't understand. So for each $$A_n$$, does it keep going until you reach $$2013$$? i.e. $$A_{2012} = {2012}^{2013}$$ and $$A_{2013} = 2013$$?

- 4 years, 2 months ago

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Yes.. This is probably because this is a sequence of integers until 2013.

- 4 years, 2 months ago

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