Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Answer
A number that leaves a remainder of 2 when divided by 3 is $=3 x+2$ A number that leaves a remainder of 3 when divided by 4 is $=4 x+3$ A number that leaves a remainder of 4 when divided by 5 is $=5 x+4$ L.C.M of 3,4 , and $5=60$ All the numbers are the same, so $4 x+3=3 x+2$ $\begin{array}{l}4 x-3 x=2-3 \\x=-1\end{array}$ Each remainder is 1 less than the divisor. Hence, the number is 1 less than the L.C.M $=(60-1)=59$. So, 59 is the smallest number that satisfies all the given conditions.
Tathagat has written several numbers that leave a remainder of 2 when divided by $6 . He$ claims, "If you add any three such numbers, the sum will always be a multiple of $6 . "$ Is Tathagat's claim true?
Answer
A number that leaves remainder of 2 when divided by 6 can be written as $6 k +2$. Three such numbers are: $(6 a+2),(6 b+2),(6 c+2)$. $(6 a+2)+(6 b+2)+(6 c+2)=6(a+b+c)+6=6(a+b+c+1)$ This sum is divisible by 6 . So yes, Tathagat's claim is always true. Example : Take $20,26,32 \rightarrow$ sum $=78 \rightarrow$ divisible by 6 . Take $2,8,14 \rightarrow$ sum $=24 \rightarrow$ divisible by 6 .
The sum of four consecutive numbers is 34. What are these numbers?
Answer
Let four consecutive numbers be $x,(x+1),(x+2)$ and $(x+3)$. $\begin{array}{l}x+(x+1)+(x+2)+(x+3)=34 \\x+x+1+x+2+x+3=34 \\4 x+6=34 \\4 x=34-6 \\4 x=28 \\x=28 / 7=7\end{array}$ $\begin{array}{l}\text { So, }(x+1)=7+1=8 \$x+2)=7+2=9 \$x+3)=7+3=10\end{array}$ Therefore, the given four consecutive numbers are $7,8,9$, and 10 .