Everyone Will Subtract the Products. You?
\large N=1434661 \times 3785648743 - 100020304 \times 54300201
Given that 0 ≤ N < 9 0 , find N .
The answer is 19.
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2 solutions
Shouldn't you prove that 0 ≤ N < 9 0 is true in the first place? How do you know that the author isn't lying?
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I've made that an explicit assumption, since I believe that there isn't a good way to show that 0 ≤ N < 9 0 < 1 0 0 .
I also left it as < 9 0 , since that somewhat hides the trick of immediately Mod 100.
Are you suggesting there's a better approach?
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No I do not. Unless you consider "elementary tedious long multiplcation" as a better approach.
The previous writeup only hinted that N satisfy this inequality, but it's not made an explicit assumption.
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@Pi Han Goh – Ah. Okay. I only saw the explicit version...
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@Paul Hindess – Oh, I just got a (only slightly) simpler solution.
Show that the last digit is 9 via mod 10.
Then use divisibility rule of 11 to show that N mod 11 = 8.
Then use divisibility rule of 3 to show that N mod 3 = 1
This leaves us with 19 only.
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@Pi Han Goh – Hm, (If I'm reading your solution correctly,) you have shown that by CRT N ≡ 1 9 ( m o d 3 3 0 ) . Why does that imply that N ≡ 1 9 ( m o d 1 0 0 ) ?
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@Calvin Lin – Sorry, let me rephrase it
We just need to show that N mod 10 = 9, and N mod 11 = 8. Using CRT, we can uniquely determine that N= 19 (given the constraint 0<= N < 90).
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@Pi Han Goh – Ah ic. Yea, it doesn't seem like there is an easy way around removing the constraint.
The last two digits of the two product are 1 0 ⋅ ( 6 × 3 + 1 × 4 ) + 1 × 3 = 2 2 3 ∼ 2 3 ; 4 × 1 = 4 ∼ 0 4 . Therefore the answer is N = 2 3 − 0 4 = 1 9 .
As we are told 0 ≤ N < 9 0 , we can work modulo 100.
This gives us 6 1 × 4 3 − 4 × 1 = 2 6 2 3 − 4 = 2 6 1 9 = 1 9 m o d 1 0 0 .
So the answer is 19.
[If we wish to be more efficient, we can ignore the 4 0 × 6 0 within the 6 1 × 4 3 calculation!]