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Number Play Examine each of the following statements, and determine whether Math it is ‛Always true’, ‛Sometimes true’, ‛Never true’. Talk 6. If a number is divisible by both 9 and 4, it must be divisible by 36. 7. If a number is divisible by both 6 and 4, it must be divisible by 24. In general, if A is divisible by k and A is also divisible by m, then A is divisible by the LCM of k and m. This is because A is a multiple of k and also a multiple of m, so A’s prime factorisation should contain the prime factorisation of LCM (k, m). 8. When you add an odd number to an even number we get a multiple of 6. We know that multiples of 6 are all even Can I write an even and an odd number as 2n and numbers. The sum of an odd number and an 2n+1 instead? even number will be an odd number. Therefore, thisstatementisnevertrue.Wecanalsoexplain this algebraically. Suppose, (2n) + (2m + 1) = 6j, where 2n is an even number, 2m + 1 is an odd number, and 6j is a multiple of 6. Then 2n + 2m = 6j – 1 2 (n + m) = 6j – 1 which means 2(n + m), which is an even number, should be equal to 6j – 1, which is an odd number. This is never true. What Remains? Find a number that has a remainder of 3 when divided by 5. Write more such numbers. Which algebraic expression(s) capture all such numbers? 3k (i) 3k + 5  (ii) 3k – 5  (iii)5   (iv) 5k + 3  (v) 5k – 2   (vi) 5k – 3 The numbers that leave a remainder of 0 when divided by 5 are the multiples of 5. But we want numbers that leave a remainder of 3 when divided by 5. These numbers are 3 more than multiples of 5. Multiples of 5 are of the form 5k. So, numbers that leave a remainder of 3 when divided by 5 are those of the form 5k + 3 5 . r . w k = 0 1 2 3 4 . s 5k + 3 = 3 8 13 18 23 121 Reprint 2026-27