Hi All,
Please help me with this problem:
What is the least number which leaves remainder 1, 2, 3 and 4 when divided successively by 2, 3, 5 and 7?
Thank you,
Ashwin Kumar D P.
Bangalore, India.
Hi All,
Please help me with this problem:
What is the least number which leaves remainder 1, 2, 3 and 4 when divided successively by 2, 3, 5 and 7?
Thank you,
Ashwin Kumar D P.
Bangalore, India.
$\displaystyle x \equiv 1 \mod 2$ Equation1
$\displaystyle x \equiv 2 \mod 3$ Equation2
$\displaystyle x \equiv 3 \mod 5$ Equation3
$\displaystyle x \equiv 4 \mod 7$ Equation4
Use the Chinese Remainder Theorem.
However, for this question try this:
The values that satisfy Eqn4 are
4, 11, 18, 25, 32, 39, ...
They are 4 + multiples of 7.
Use those values in Eqn3
$\displaystyle 4 \equiv 4 \mod 5$
$\displaystyle 11 \equiv 1 \mod 5$
$\displaystyle 18 \equiv 3 \mod 5$
18 satisfies Eqn4 & Eqn3
Use 18 + multiples of 35 ( 7 times 5 ) for Equation2
18, 53, 88, 123, ...
Find the one that satisfies equation2.
Then a similar procedure for Equation1
.