Question 5 - Figure it out - Page 122, 123 - Chapter 5 Class 8 - Number Play (Ganita Prakash)

Transcript

Question 5 “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”Let the Number of pebbles = N Let’s read it one by one “When I group them in 3’s, one stays with me” So, Number is a multiple of 3 plus 1 Multiple of 3 = 3, 6, 9, 12, 15, 18, …. Number can be = 3 + 1, 6 + 1, 9 + 1, 12 + 1, 15 + 1,… = 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97, 100, … “Try pairing them up — it simply won’t do” The number is Odd. Number can be = 7, 13, 19, …. “Group them by 5, yet one’s still around The number must end in a 1 or 6. Since it's odd, it must end in 1. Number can be = 31, 61, 91 “But grouping by seven, perfection is found.” The number must be divisible by 7 exactly. Check candidates: Is 31 divisible by 7? No. Is 61 divisible by 7? No. Is 91 divisible by 7? Yes! (13 × 7 = 91). “More than one hundred would be far too bold” The number is less than 100 Thus, the number is 91