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發問:
H.C.F: 1164, 1455, 1746 H.C.F: 354, 424, 459 There is a number which ies greater than 1000. When it is divided by 12, 14 and 21, it will give 3, 5, and 12 as a remainder respectively. Which is the least possible value of this number?
最佳解答:
Question 1 ２｜１１６４ └－－－－－ ２｜５８２ └－－－－－ ３｜２９１ └－－－－－ ９７ ３｜１４５５ └－－－－－ ５｜４８５ └－－－－－ ９７ ２｜１７４６ └－－－－－ ３｜８７３ └－－－－－ ３｜２９１ └－－－－－ ９７ 1164 = 22 × 3 × 97 1455 = 3 × 5 × 97 1746 = 2 × 32 × 97 HCF = 3 × 97 = 291 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Question 2 ２｜３５４ └－－－－ ３｜１７７ └－－－－ ５９ ２｜４２４ └－－－－ ２｜２１２ └－－－－ ２｜１０６ └－－－－ ５３ ３｜４５９ └－－－－ ３｜１５３ └－－－－ ３｜５１ └－－－－ １７ 354 = 2 × 3 × 59 424 = 23 × 53 459 = 33 × 17 HCF = 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Question 3 Let x be the number. x = 12A + 3 = 12A' - 9 where integer A' = A + 1 x = 14B + 5 = 14B' - 9 where integer B' = B + 1 x = 21C + 12 = 21C' - 9 where integer C' = C + 1 12 = 22 × 3 14 = 2× 7 21 = 3× 7 L.C.M. of 12, 14, 21 is 22 × 3 × 7 = 84 For any integer D, the number (84D - 9) satisfies the three conditions. Consider 1000 ÷ 84 = 11 ... 76 Put D = 12, 84D - 9 = 999 < 1000 Put D = 13, 84D - 9 = 1083 > 1000 The least possible value of the number is 1083.
其他解答:960E6207E048F01D
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f.1 MATHS發問:
H.C.F: 1164, 1455, 1746 H.C.F: 354, 424, 459 There is a number which ies greater than 1000. When it is divided by 12, 14 and 21, it will give 3, 5, and 12 as a remainder respectively. Which is the least possible value of this number?
最佳解答:
Question 1 ２｜１１６４ └－－－－－ ２｜５８２ └－－－－－ ３｜２９１ └－－－－－ ９７ ３｜１４５５ └－－－－－ ５｜４８５ └－－－－－ ９７ ２｜１７４６ └－－－－－ ３｜８７３ └－－－－－ ３｜２９１ └－－－－－ ９７ 1164 = 22 × 3 × 97 1455 = 3 × 5 × 97 1746 = 2 × 32 × 97 HCF = 3 × 97 = 291 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Question 2 ２｜３５４ └－－－－ ３｜１７７ └－－－－ ５９ ２｜４２４ └－－－－ ２｜２１２ └－－－－ ２｜１０６ └－－－－ ５３ ３｜４５９ └－－－－ ３｜１５３ └－－－－ ３｜５１ └－－－－ １７ 354 = 2 × 3 × 59 424 = 23 × 53 459 = 33 × 17 HCF = 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Question 3 Let x be the number. x = 12A + 3 = 12A' - 9 where integer A' = A + 1 x = 14B + 5 = 14B' - 9 where integer B' = B + 1 x = 21C + 12 = 21C' - 9 where integer C' = C + 1 12 = 22 × 3 14 = 2× 7 21 = 3× 7 L.C.M. of 12, 14, 21 is 22 × 3 × 7 = 84 For any integer D, the number (84D - 9) satisfies the three conditions. Consider 1000 ÷ 84 = 11 ... 76 Put D = 12, 84D - 9 = 999 < 1000 Put D = 13, 84D - 9 = 1083 > 1000 The least possible value of the number is 1083.
其他解答:960E6207E048F01D
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