Find the smallest number which gives reminder 1, 2 and 3 when divided by 7, 9 and 11 respectively? Answer: 344 Let the number be p P/7 = remainder 1 P/9 = remainder 2 P/11 = remainder 3 and 2P/7 = remainder 2 2P/9 = remainder 4 2P/11 = remainder 6 Now you can see the pattern Notice that if we add 5 to 2P , these equation will become completely divisible by 7,9 and 11 respectively, i.e. remainder becomes 0. So, 2P+5/7 = remainder 0 2P+5/9 = remainder 0 2P+5/11 = remainder 0 this means 2P+5 is divisible by 7,9 & 11. So it will be divisible by 7x9x11=693 2P+5 = 693 2P = 688 P = 344 Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.