Help with maths question on exponents please

[InterruptingCoww]

Hello,
How on earth do I solve this? Please help, it’s on a primary school maths comp but it is the last question so think it’s more high school level. Many thanks for any help

Sorry, meant to say calculators are not allowed!

The answer is 43.

The answer comes up on a Google search of the question. Neither my laptop calculator of my phone calculate the answer in that form. I have an engineering degree, a primary teaching degree, but I can't explain at primary school level. I can understand why it's that answer, but it's beyond what I'd normally teach to year 6.

7 last digit 7
7 x 7 = 49 last digit 9
7 x 7 x 7 - the last digit will be 3 because 9 x 7 = 63
7 x 7 x 7 x 7 - the last digit will be 1 because 3 x 7 = 21

So the pattern for last digits is
7 9 3 1

So for 7^77 the last digit will be 3

Thanks both, the correct answer is 7, according to the answer key. We did workout the pattern 7 9 3 1 and then 19 x 4 sets of the pattern brings us up to the 76th number so the the 77th number would be a 7. We did it with a calculator though so was wondering how we could have done this without one. We were completely stumped! Thanks both for answering.

I think it's 7 because 7 to the power of 77 = 777 = 118181 3865805958 7997686841 4312001964 4340385488 3676992345 8287039207 and the internet tells me the last 2 digits are the remainder. exponentiations.com/7-to-the-77th-power and www.omnicalculator.com/math/remainder
And the first link mentions the inverse of 7^77 is 7. Is it a prime number trick thing? Of course this might be wrong because I'm up at 4am searching maths problems so not a reliable source!

the number 7⁷⁷ is a 65 digit number, the child is not supposed to calculate it.
this is certainly well beyond primary level and may have been set as a way to find out which pupils are using the internet to help rather than just doing the work to the best of their understanding

but as @marcopront points out, there is a pattern.

whenever any number that ends in 7 is multiplied by 7, the answer will end in a 9.

whenever any number that ends in 9 is multiplied by 7, the answer will end in a 3.

whenever any number that ends in 3 is multiplied by 7, the answer will end in a 1.

whenever any number that ends in 1 is multiplied by 7, the answer will end in a 7.

if a child can spot this pattern then they can deduce that the value of 7⁷⁷ will end in a 7, because 7⁷⁶ will have ended in a 1 because 76 is a multiple of 4.

that doesn't prove that the remainder when divided by 100 would be 7, but 7 is the only compatible answer among the choices given

Sorry, something went wrong with my copy paste there and 777 should be 7 superscript 77 but anyway I see you found the answer so that's good. I can rest easy now... 😁

Sorry the last digit will be a 7
77 = 19 x 4 + 1

Thanks all, we have just been down a rabbit hole multiplying other numbers to find patterns. 😂

The patterns are interesting

Ludicrously difficult for that age, I wouldn't give it a moment's thought.

I think the key to this at primary level is to recognise that the remainder when any number that is a round hundred is divided by 100 is zero.

So to answer the question you only need to look at the pattern formed by the last two digits. It doesn't matter what the hundreds digit is, when you multiply it by 7 it will still be divisible by 100.

I don't have a pen and paper to hand so I haven't worked this through, but you could set up the first few in the same way as marcopront says (but taking the last two digits not just the last one), look for a pattern, and when you see it use FaazoHuyzeoSix's reasoning to see where the 77th one falls in that pattern. Bingo.

yes I guess the learning outcome is to make sure kids realise that all these kinds of multiplications have a pattern that repeats on up to infinity so that you can always deduce the last digit of a very large power calculation despite not knowing the whole value.
when multiplying repeatedly
the pattern for 2 is 2,4,8,6
for 3 it will be 3,9,7,1
for 4 it will be 4,6,4,6
for 5, all powers will end in 5
for 6, all powers will end in 6
we've done 7
for 8, it will be 8,4,2,6
for 9 it will be 9,1,9,1

what I don't quite get is why the question seems to imply that it's also possible to deduce the penultimate digit without knowing the full sum.

i can deduce that 3²⁵ must end in 3 because 24 is a multiple of 4 so 3²⁴ must end in a 1 so 3²⁵ has to end in 3. But is it possible to deduce without doing the calculation whether it ends in 23,43,63 or what?

@FaazoHuyzeoSix

what I don't quite get is why the question seems to imply that it's also possible to deduce the penultimate digit without knowing the full sum.

The pattern for 6 and 7 is actually for the last two digits.
I have just discovered that.

It is also interesting looking at the last digit of powers.
So powers of 2 end in
1, 4, 9, 6, 5, 6, 9, 4, 1
Powers of 5 end in
1, 2, 3, 4, 5, 6, 7, 8, 9

Answer already been fairly well explained here.

For those questioning the difficulty - the format of this is for a Maths Challenge questions. It’s meant to be difficult, it’s to stretch children beyond the curriculum. I’m not familiar with the primary maths challenge but this is certainly along the level that would come up at junior maths challenge level.

To give an idea of how difficult this actually is in my Cambridge interview for maths the first question was:-

2^1996 - what is the first digit, the last digit and how many digits are there?

(This was in 1996.)

Exactly this. That's why the question asks for the remainder when divided by 100. Anything in the 100s column or above will have zero remainder. So you only need the last two digits to answer the question.

This is a very difficult question but at least it makes sense of some of the homework my ds recently got. In year 1 they were always asked to explain the patterns of each of their timetables up to 100x and then predict the outcome of any x times y question on the basis of the pattern. I am still not sure what the real life use is, but I guess some schools hope to do better in such competitions than others 😀

Quite often the competitions do want you to look at the answers and work out which is the only compatible one for some reason, without working out the answer fully. So that's probably want they wanted here, knowing the last digit is 7 even if you don't know what the second last digit is.

If it's the primary maths challenge, there should be an answer page on the site

Ooh, that's hard.
In 1984 I had to prove the test for visibility by 3, 9 and 11.

I also had to explain what happens if you balance a ruler by its ends on two fingers and then gradually move your fingers together.

Helpful for when you need to factorise a number (eg simplifying fractions, or solving quadratics in secondary), and also for learning an 'instinct' as to whether your answer 'looks right'.

7^1 ends in 07
7^2 ends in 49
7^3 = 343, ends in 43
7^4 = 2401, ends in 01.
the 01 ending means that if you multiply the 7s in groups of 4, the answer will end in 01.
777= 772 x 7^3
so it will end in 01 x 43 = 43
this method is quite difficult.

an easier method is only looking at the last digit and spot the pattern 7,9,3,1 and notice that the answer ends in 3. Out of all the options, only 43 ends in 3 so the answer is C.

the question in the original post is all about spotting pattern. A good y8-y9 should be able to do that.

The interview question is much harder. I think you will need to use log to put a lower bound and UB using base/power of 10 to find out how many digits there are and what the first digit is. The last digit bit is trivial of course.

The last digit bit is trivial of course.

Oh that wonderful maths word 'trivial', often meaning anything but as an undergraduate.