Probability problem

1a2s3d4f5g6h7j8k9l

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Hi people!

Find the smallest positive integer, N, satisfying the following conditions:
If N is divided by 7, the remainder is 4.
If N is divided by 8, the remainder is 5.
If N is divided by 9, the remainder is 6.


I need help starting on this problem... please help!

Thank you.
 
I think this has more to do with multiples and factors.

Try find out the minimum value of N satisfying the first condition only:

\(\displaystyle N = \frac{7x + 4}{7}\)

And x is a whole number. Smallest value of x is 1, solve for N.

You can even use an arithmetic progression to get all the numbers that satisfy the equation.

Then, use the second condition:

\(\displaystyle N = \frac{8x + 5}{8}\)

Find another value of N for x = 1.

Construct another arithmetic progression.

Do the same for the last one.

This will give you an idea of what number you're looking for.
 
Hello, 1a2s3d4f5g6h7j8k9l

Find the smallest positive integer \(\displaystyle N\) satisfying the following:
. . If \(\displaystyle N\) is divided by 7, the remainder is 4.
. . If \(\displaystyle N\) is divided by 8, the remainder is 5.
. . If \(\displaystyle N\) is divided by 9, the remainder is 6.

From the given statements, we have:

. . \(\displaystyle \begin{Bmatrix}N &=& 7a + 4 & [1] \\ N &=& 8b + 5 & [2] \\ N &=& 9c + 6 & [3] \end{Bmatrix}\;\text{ for integers }a,b,c \ge 0\)


Equate [1] and [2]: .\(\displaystyle 7a + 4 \:=\:8b + 5 \quad\Rightarrow\quad 7a \:=\:8b + 1 \)

. . . . . . . . . . . . . . . \(\displaystyle a \:=\:\dfrac{8b+1}{7} \quad\Rightarrow\quad a \:=\:b + \dfrac{b+1}{7}\)

Since \(\displaystyle a\) is an integer, \(\displaystyle b+1\) is a multiple of 7.
. . \(\displaystyle b+1 \,=\,7p \quad\Rightarrow\quad b \,=\,7p-1\) .(a)


Equate [2] and [3]: .\(\displaystyle 8b + 5 \:=\:9c + 6 \quad\Rightarrow\quad 8b \:=\:9c+1\)

Substitute (a): .. \(\displaystyle 8(7p-1) \:=\:9c + 1 \quad\Rightarrow\quad 56p - 8 \:=\:9c+1\)

. . . . . . . . . . . . . . . .. \(\displaystyle 56p \:=\:9c + 9 \quad\Rightarrow\quad p \:=\:\dfrac{9(c+1)}{56}\)

Since \(\displaystyle p\) is an integer, \(\displaystyle c+1\) is a multiple of 56.
. . For minimum values, \(\displaystyle c = 55 \quad\Rightarrow\quad p \,=\,9\)

Substitute into (a): .\(\displaystyle b \:=\:7(9)-1 \quad\Rightarrow\quad b \:=\:62\)

Substitute into \(\displaystyle [2]\!:\;N \:=\:8(62) + 5 \quad\Rightarrow\quad \boxed{N \:=\:501}\)
 
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