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secondary 3 | A Maths
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Hi, I am stuck with these questions. Can't solve any of them. Kindly advise the working solutions. Thanks a lot!
Date Posted: 4 years ago
Views: 491
Good afternoon Candice! This question is probably advanced for me, but I will have a look at it at a later time and attempt them within my ability.
For the first one, it’s very straightforward. Factorials (with exclamation sign) means the product of all whole numbers up to that number, so 5! means 1 x 2 x 3 x 4 x 5. This is 120. Multiplying by future numbers do not change the last digit “0”, so 2022! will have a last digit of zero.
If I am able to come up with a solution, I will post them.
For the first one, it’s very straightforward. Factorials (with exclamation sign) means the product of all whole numbers up to that number, so 5! means 1 x 2 x 3 x 4 x 5. This is 120. Multiplying by future numbers do not change the last digit “0”, so 2022! will have a last digit of zero.
If I am able to come up with a solution, I will post them.
Part two I believe has to do with the fact that every “5” that is present will contribute to the last digit being zero, because 2 and 5 multiply to 10 which obviously adds a zero to the back. There are more “2”s than “5”s in the list, so if we can find the number of “5”s contained in 2023!, we should be able to obtain the number of zeroes.
In short, we have to find
1. Numbers which are multiples of 625; these will contain four “5”s per number
2. Numbers which are multiples of 125 but not 625; these will contain three “5”s per number
3. Numbers which are multiples of 25 but not 125; these will contain two “5”s per number
4. Finally, numbers which are multiples of 5 but not 25; these will contain one “5” per number
For the 2s, there are simply way too many of them, because within the first 10 numbers, 2, 4, 6, 8 and 10 already contain eight “2”s within them as opposed to only two “5”s and there will be even more “2”s within the first 100 numbers.
In short, we have to find
1. Numbers which are multiples of 625; these will contain four “5”s per number
2. Numbers which are multiples of 125 but not 625; these will contain three “5”s per number
3. Numbers which are multiples of 25 but not 125; these will contain two “5”s per number
4. Finally, numbers which are multiples of 5 but not 25; these will contain one “5” per number
For the 2s, there are simply way too many of them, because within the first 10 numbers, 2, 4, 6, 8 and 10 already contain eight “2”s within them as opposed to only two “5”s and there will be even more “2”s within the first 100 numbers.
Sorry, part iii is actually straightforward because the last digit of 5! onwards is zero, so last digit of the sum
= last digit of (1! + 2! + 3! + 4!) (because the rest have last digits zero and will not affect this result)
= last digit of (1 + 2 + 6 + 24)
= last digit of (33)
= 3
= last digit of (1! + 2! + 3! + 4!) (because the rest have last digits zero and will not affect this result)
= last digit of (1 + 2 + 6 + 24)
= last digit of (33)
= 3
See 3 Answers
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Good afternoon Candice! Here are my workings for parts i and iii.
Date Posted:
4 years ago
Great explanation, Mr Eric. I understand your working instantly. They are super clear :) Thanks!
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Good afternoon Candice! Here are my workings for part ii. We note that for every one “2” paired with one “5”, we obtain one “10”. Note that the number of “10”s which we can obtain is limited to the available number of “2”s or “5”s. So for example, 2^4 * 5 = 80 will only contain one “10” because there is only one “5” (and therefore no more “10”s can be generated).
As such, I found that there are 503 “5”s in the list altogether, versus much more than 1011 “2”s in the list. Since the number of “5”s is very limited, there can only be a series of 503 consecutive zeroes at the end of the list.
As such, I found that there are 503 “5”s in the list altogether, versus much more than 1011 “2”s in the list. Since the number of “5”s is very limited, there can only be a series of 503 consecutive zeroes at the end of the list.
Date Posted:
4 years ago
As always very detailed awesome explanation :) Thumbs up! Thanks Mr Eric!
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Hi Candice, my answer is different from the one posted by Eric, maybe you can refer to the answer key.
Keys to my working are:
1. For every 5 incremental of factorial, you get additional zero because the whole sum is multiply by 10 or having 5 multiply by any past even number.
2. Then we need to consider when the number reach in the hundreds as at 100,200,300, etc, instead of multiply by 10, the whole number is multiply by 100. So there is additional zero introduced as compared to point no1
3. Same thing with thousands, instead of multiplying by 100, we multiply by 1000. Again there is introduction of additional zero.
Keys to my working are:
1. For every 5 incremental of factorial, you get additional zero because the whole sum is multiply by 10 or having 5 multiply by any past even number.
2. Then we need to consider when the number reach in the hundreds as at 100,200,300, etc, instead of multiply by 10, the whole number is multiply by 100. So there is additional zero introduced as compared to point no1
3. Same thing with thousands, instead of multiplying by 100, we multiply by 1000. Again there is introduction of additional zero.
Date Posted:
4 years ago
Your answer looks very logical. Maybe my answer is incorrect. But I am wondering what happens upon reaching 25! in your list, because apparently the multiplication of 25 introduces two zeroes (I think).
Say 24 * 25 = 600. Two zeroes are instantly introduced provided there are sufficient “2”s.
Say 24 * 25 = 600. Two zeroes are instantly introduced provided there are sufficient “2”s.
Thanks for your advice, Mr Ricky. I appreciate your time spent working on my question. Mr Eric's answer is correct, its 503.
Ah yes I forgot to consider that. Thanks for the enlightenment
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