Problem Solving

find the units digits of

7^11^13^16^19^21

is the ans "1"

I think the answer is 1

Can rearrange the eqn to 7^(11*13*16*19*21)

Units digit of 7^n will repeat 7,9,3,1 with 7 or 3 on odd powers and 9 or 1 on even powers.

We know it will be an even power since 16 is a factor (can narrow the ans down to 9 or 1)

Units digit should be 1 since the factor 16 tells us the power is a multiple of 4, and multiples of 4 are the largest powers that gives us a unique units digits (in this case 1).

I have a doubt, I agree with canman that the expression can be writter as

7^(11*13*16*19*21)

Now looking at the power in the expression, the power can be expressed
4N were N=11*13*4*19*21

hence we have 7^(4N)...so the units digit should be 1

reachac wrote:I have a doubt, I agree with canman that the expression can be writter as

7^(11*13*16*19*21)

Now looking at the power in the expression, the power can be expressed
4N were N=11*13*4*19*21

hence we have 7^(4N)...so the units digit should be 1

I definitely dont think you can do that.

e.g. what is 2^3^2 are you saying that 2^6 = 2^9?

Hope its clear...

Agree with parallel chase...

I see your point- I applied a rule that required parantheses around each of the terms. Sorry for any confusion. Back to the starting line for me.

I find 1

parallel_chase wrote:

reachac wrote:I have a doubt, I agree with canman that the expression can be writter as

7^(11*13*16*19*21)

Now looking at the power in the expression, the power can be expressed
4N were N=11*13*4*19*21

hence we have 7^(4N)...so the units digit should be 1

I definitely dont think you can do that.

e.g. what is 2^3^2 are you saying that 2^6 = 2^9?

Hope its clear...

Actually, it's a great solution.

Unless I'm mistaken, reachac was merely saying that we can rewrite the exponent as a multiple of 4.

For example, we could rewrite 2^20 as 2^(4*5).

Reachac was NOT saying that we can rewrite 2^20 as 2^4 * 2^5, which would in fact be 2^9.

So, since we know that the exponent is a multiple of 4, and we know that the units digit of powers of 7 is a 4-number repeating cycle (7, 9, 3, 1), we know that the units digit of the number in the question will definitely be 1.

Best way is described by Stuart, here is method n°)2

7^1: unit 7
7^2: unit 9
7^3: unit 3
7^4: unit 1

So 7^11 has unit 3

Now let's see the unit digit of 3^13
3^1: unit 3
3^2: unit 9
3^3: unit 7
3^4: unit 1

So 3^13 has unit 3
And 3^16 has unit 1

And 1^19^21 has unit 1

I think I made a mistake by taking the exopnent powers as the power of the whole expression, indeed it look now that each exponent(barring 11) is the power of the precceding exponent.

If thats the case, the answer is 3

If you solve for the powers it will be always a multiple of 11, therefore I truly think 7^11 has units digit as 3 and same for every other multiple of 11

Let me know what you guys think.

parallel_chase wrote:If you solve for the powers it will be always a multiple of 11, therefore I truly think 7^11 has units digit as 3 and same for every other multiple of 11

Let me know what you guys think.

I am really sorry but I don agree with ur method, to exemplify,

11 = 4*2)+3 -->4n+3 form
11^2=121 = (4*30)+1 --> 4n+1 form

for these two forms, the ans (remainder for this question) would be different i.e 3 in case of the 4n+3 formand 7 in case of the 4n+1 form

hence we need to determine the power of 11 as odd or even

Totally agree

Thanks a ton!!!!