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secondary 4 | A Maths
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Date Posted: 3 years ago
Views: 248
Here's an alternative way without using modular arithmetic.
The idea is to look at the last 3 digits since the fourth last digit and beyond will be multiples of 1000.
Using a simpler example first,
9 × 99 = 891
If divided by 1000, the remainder is 891 since we can't divide at all.
Next,
9 × 99 × 999
= 891 × (1000 - 1)
= 891 × 1000 - 891
= 891 000 - 891
= 890 109
= 890 000 + 109
= 890 × 1000 + 109
891 000 has last 3 digits being zeroes.
So if we subtract 891, the last 3 digits are 109.
When 890 109 is divided by 1000, the remainder is 109 since that 890 000 is a multiple of 1000 and is divisible with no remainder.
Likewise,
9 × 99 × 999 × 9999
= 890 109 × (10000 - 1)
= 890 109 × 10000 - 890 109
= 8 901 090 000 - 890 109
= 8 900 199 891
= 8 900 199 000 + 891
= 8 900 199 × 1000 + 891
As we can see, the last 3 digits of 8 901 090 000 are zeroes also.
Subtracting that 109 gives you 891 in the last 3 digits.
The remainder is 891.
The idea is to look at the last 3 digits since the fourth last digit and beyond will be multiples of 1000.
Using a simpler example first,
9 × 99 = 891
If divided by 1000, the remainder is 891 since we can't divide at all.
Next,
9 × 99 × 999
= 891 × (1000 - 1)
= 891 × 1000 - 891
= 891 000 - 891
= 890 109
= 890 000 + 109
= 890 × 1000 + 109
891 000 has last 3 digits being zeroes.
So if we subtract 891, the last 3 digits are 109.
When 890 109 is divided by 1000, the remainder is 109 since that 890 000 is a multiple of 1000 and is divisible with no remainder.
Likewise,
9 × 99 × 999 × 9999
= 890 109 × (10000 - 1)
= 890 109 × 10000 - 890 109
= 8 901 090 000 - 890 109
= 8 900 199 891
= 8 900 199 000 + 891
= 8 900 199 × 1000 + 891
As we can see, the last 3 digits of 8 901 090 000 are zeroes also.
Subtracting that 109 gives you 891 in the last 3 digits.
The remainder is 891.
What we realise is that the remainder alternates between 891 and 109 as you increase the number of terms.
When there are an even number of terms, (eg. there are two terms 9 and 99 in 9 × 99), the remainder is 891.
When there are an odd number of terms, the remainder is 109.
Since the question has the first term being 9 (one 9) and the last term having 999 '9's , there are 999 terms altogether.
999 is odd. So the remainder must be 109.
When there are an even number of terms, (eg. there are two terms 9 and 99 in 9 × 99), the remainder is 891.
When there are an odd number of terms, the remainder is 109.
Since the question has the first term being 9 (one 9) and the last term having 999 '9's , there are 999 terms altogether.
999 is odd. So the remainder must be 109.
To represent it in a single working ,
9 × 99 × 999 × 9999 × 99999 × ...
= 891 × (1000 - 1) × (10000 - 1) × (100000 - 1) × ...
We only have to be concerned with the multiplication of 891 with every -1 you see in each brackets.
An odd number of terms gives -891, resulting in 109 for the last 3 digits.
An even number of terms gives +891, resulting in 891 for the last 3 digits.
All other products will give multiples of 1000 and result in the last 3 digits being 000. So we don't have to look at them.
This can be summarised as :
Remainder = 500 + 391(-1)ⁿ
where n = number of '9's in the last term, n ≥ 2
There are 999 terms, and 999 is odd.
So 109 is the remainder.
9 × 99 × 999 × 9999 × 99999 × ...
= 891 × (1000 - 1) × (10000 - 1) × (100000 - 1) × ...
We only have to be concerned with the multiplication of 891 with every -1 you see in each brackets.
An odd number of terms gives -891, resulting in 109 for the last 3 digits.
An even number of terms gives +891, resulting in 891 for the last 3 digits.
All other products will give multiples of 1000 and result in the last 3 digits being 000. So we don't have to look at them.
This can be summarised as :
Remainder = 500 + 391(-1)ⁿ
where n = number of '9's in the last term, n ≥ 2
There are 999 terms, and 999 is odd.
So 109 is the remainder.
See 2 Answers
done
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clear
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Problem is trivial once you learn modular arithmetics
Date Posted:
3 years ago
Last sentence should be -891. Forgot the - sign
For the second line, the last term's power should be 999 rather than 1000.
Also, strictly speaking ,
999 mod 1000 = 999
9999 mod 1000 = 9999, and not -1.
Perhaps you meant:
999 ≡ -1 (mod 1000) and 9999 ≡ -1 (mod 1000)
@Nancy : Alternative style of presentation in the last two steps
Based on the multiplication property:
If a ≡ b (mod m) and c ≡ d (mod m), then ac = bd (mod m)
Then 999 ≡ 9999 ≡ 99999 ≡ ... ≡ (999 '9's) ≡ -1 (mod 1000)
Then 999 × 9999 × 99999 × ... × (999 '9's) ≡ (-1)(-1)(-1)... (mod 1000) = (-1)^997 (mod 1000) = -1 (mod 1000)
And since 891 ≡ -109 (mod 1000),
Then 9 × 99 × 999... × (999 '9's)
= 891 × 999... × (999 '9's)
≡ -109 × (-1) (mod 1000) = 109 (mod 1000)
999 mod 1000 = 999
9999 mod 1000 = 9999, and not -1.
Perhaps you meant:
999 ≡ -1 (mod 1000) and 9999 ≡ -1 (mod 1000)
@Nancy : Alternative style of presentation in the last two steps
Based on the multiplication property:
If a ≡ b (mod m) and c ≡ d (mod m), then ac = bd (mod m)
Then 999 ≡ 9999 ≡ 99999 ≡ ... ≡ (999 '9's) ≡ -1 (mod 1000)
Then 999 × 9999 × 99999 × ... × (999 '9's) ≡ (-1)(-1)(-1)... (mod 1000) = (-1)^997 (mod 1000) = -1 (mod 1000)
And since 891 ≡ -109 (mod 1000),
Then 9 × 99 × 999... × (999 '9's)
= 891 × 999... × (999 '9's)
≡ -109 × (-1) (mod 1000) = 109 (mod 1000)
See main comments for alternative method without using modular arithmetic
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