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A question about the natural number N:?
n=162a8b is a number of 6 digits, where 0=<a,b<=b. Find all a & b
such that n can be properly divided by 7.
Correction: 0=<a,b<=9
4 Answers
- PuzzlingLv 72 months agoFavourite answer
First, where you said 0 ≤ a,b ≤ *b* I assume you meant 0 ≤ a,b ≤ 9.
There are several ways to approach this. Here's just one method.
We can think of N as:
162080 + 100a + b
When we divide 162080 by 7, the remainder is 2. So the following expression will give the same remainder when divided by 7.
100a + b + 2
Also, since 98 is a multiple of 7, we can drop 98a. This will also give the same remainder when divided by 7.
2a + b + 2
Now we can easily plug in a = 0 to 9 and calculate the possible values of b where the remainder is 0.
a = 0 --> b = 5
a = 1 --> b = 3
a = 2 --> b = 1 or 8
a = 3 --> b = 6
a = 4 --> b = 4
a = 5 --> b = 2 or 9
a = 6 --> b = 0 or 7
a = 7 --> b = 5
a = 8 --> b = 3
a = 9 --> b = 1 or 8
Answers:
162085
162183
162281
162288
162386
162484
162582
162589
162680
162687
162785
162883
162981
162988
P.S. Another way is brute force all 100 choices of a (0-9) and b(0-9) and test them for divisibility by 7. I did this also and it concurs with my answers above.
- PinkgreenLv 72 months ago
n=162a8b=162080+100a+b, 0=<a,b<=9
If b is removed, then n->n'=162a8. Let
16208+10a-2b=0(mod 7), then
10a=(2b-16208)(mod 7)-----(1)
Now try b=0,1,2,...,9 in turn in (1) & find "a" accordingly.
The results are as listed:
| b |-----0-----|-----1-----|-----1-----|-----2-----|-----3-----|-----3-----|
| a |-----6-----|-----2-----|-----9-----|-----5-----|-----1-----|-----8-----|
| n | 162680 | 162281 | 162981 | 162582 | 162183 | 165883 |
| b |-----4-----|-----5-----|-----5-----|-----6-----|-----7-----|-----8-----|
| a |-----4-----|-----0-----|-----7-----|-----3-----|-----6-----|-----2-----|
| n | 162484 | 162085 | 162785 | 162386 | 162687 | 162288 |
| b |-----8-----|-----9-----|
| a |-----9-----|-----5-----|
| n | 162988 | 162589 |
For examples,
(i) From (1), input b=3, then
10a=-16202(mod 7)=>
10a=3(mod 7)=>
a=1(mod 7)=>
a=1+7k=>
a=1, k=0
a=8, k=1
=>
n=162183 or n=165883
(ii) input b=5, then
10a=-16198(mod 7)=>
10a=0(mod 7)=>
a=0+7k=>
a=0, k=0
a=7, k=1
=>
n=162085 or n=162785
etc.
If using the method of "trial & error", we
need to try at least 10*10=100 times.
- az_lenderLv 72 months ago
You mistyped this, where you wrote b<=b. Maybe you meant b<=9 ?
This is a pretty simple problem, you can start at 162085 (which is a multiple of 7) and just keep adding either 7 or 98, depending on whether adding the 7 makes the "8" turn into a "9".
For instance 162085 + 7 is too much, so you go next to
162085 + 98 = 162183.
Then 162183 + 7 is too much, so you next go to
162183 + 98 = 162281.
Then 162281 + 7 is okay, your next success is
162288.
Then 162288 + 98 = 162386.
et cetera!
Pretty soon you get above 163000 so you have to stop.
- ?Lv 72 months ago
Every seven numbers, one of them is divisible by 7. So consider
162080.
Which is the first number divisible by 7? [ you only have the seven first to try].
Answer that, then we will take it from there...