Question 1135923: A teacher wants her students to guess the three-digit number that she
is thinking. She gives these clues:
• It is divisible by both 3 and 11.
• If you subtract one, the result is divisible by both 2 and 7.
Which number is it?
Found 3 solutions by ankor@dixie-net.com, MathTherapy, greenestamps:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A teacher wants her students to guess the three-digit number that she is thinking.
She gives these clues:
• It is divisible by both 3 and 11.
It must be a multiple of 33
:
• If you subtract one, the result is divisible by both 2 and 7.
Which number is it?
:
Lets try an equation on our Ti83, check the table for integer solution
y = 
y = 40 when x = 17
Then 17*33 = 561 is the number
Subtract 1 and it is indeed divisible by 2 and 7
Answer by MathTherapy(10552)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
One method for solving the problem using formal mathematics is with modular arithmetic. The given problem says we are looking for a 3-digit number that is a multiple of 33 and is 1 more than a multiple of 14. So we need to add a bunch of 33s and divide the total by 14 to see if it is 1 more than a multiple of 14 -- i.e.,we want the remainder when divided by 14 to be 1.
With modular arithmetic, this kind of thing is made easier because we can work with only the remainders instead of the whole large numbers. Here is what the work might look like.
To start with, we note that 33 divided by 14 is 2 with remainder 5. So instead of adding a bunch of 33s, we can add 5's instead. Here we go....
n remainder
33n / 14
---------------
1 5
2 5+5 = 10
3 10+5 = 15 = 1 (15 divided by 14 leaves remainder 1; so we can use remainder 1 instead of 15)
(This is the remainder we wanted; however, 33*3 = 99; and we need a 3=digit number.)
(So we need to keep going)
4 1+5 = 6
5 6+5 = 11
6 11+5 = 16 = 2
7 2+5 =7
8 7+5 = 12
9 12+5 = 17 = 3
10 3+5 =8
11 8+5 = 13
12 13+5 = 18 = 4
13 4+5 = 9
14 9+5 = 14 = 0
15 0+5 = 5
16 5+5 = 10
17 10+5 = 15 = 1 (This is the remainder we want)
n=17 makes 33n have a remainder of 1 when divided by 14, so that's the number we are looking for.
33*17 = 561; and 561-1 = 560 = 14*40
Note: A student familiar with modular arithmetic would have found the solution with much less work. Having found that n=3 gives a solution that is too small (99, a 2-digit number; where a 3-digit number is required), he would know that, because 14 and 33 have no common factor, the next multiple of 33 that would also have a remainder of 1 when divided by 14 is going to be with n = 3+14 = 17.
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