第 10 类 NCERT 解决方案 - 第 1 章实数 - 练习 1.2
问题 1：将每个数字表示为其质因数的乘积：
(一) 140
(二) 156
(三) 3825
(四) 5005
(五) 7429
解决方案：
(i) 140
Taking the LCM of 140,
140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7
Therefore, the product of the prime factors is 22 × 5 × 7
(ii) 156
Taking the LCM of 156,
156 = 2 × 2 × 3 × 13 × 1 = 22 × 3 × 13
Therefore, the product of the prime factors is 22 × 3 × 13
(iii) 3825
Taking the LCM of 3825,
3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32 × 52 × 17
Therefore, the product of the prime factors is 32 × 52 × 17
(iv) 5005
Taking the LCM of 5005,
5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13
Therefore, the product of the prime factors is 5 × 7 × 11 × 13
(v) 7429
Taking the LCM of 7429,
7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23
Therefore, the product of the prime factors is 17 × 19 × 23
问题 2：找出以下整数对的 LCM 和 HCF，并验证 LCM × HCF = 两个数的乘积。
(一) 26 和 91
(ii) 510 和 92
(iii) 336 和 54
解决方案：
(i) 26 and 91
Taking the LCM of 26,
26 = 2 × 13 × 1
Taking the LCM of 91,
91 = 7 × 13 × 1
Therefore, LCM of 26 and 91 together = 2 × 7 × 13 × 1 = 182
HCF of 26 and 91 = 13
Now, the product of 26 and 91 = 26 × 91 = 2366
And the product of LCM and HCF = 182 × 13 = 2366
Therefore, LCM × HCF = product of the 26 and 91.
(ii) 510 and 92
Taking the LCM of 510,
510 = 2 × 3 × 17 × 5 × 1
Taking the LCM of 92,
92 = 2 × 2 × 23 × 1
Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
HCF of 510 and 92 = 2
Now, the product of 510 and 92 = 510 × 92 = 46920
And the product of LCM and HCF = 23460 × 2 = 46920
Therefore, LCM × HCF = product of the 510 and 92.
(iii) 336 and 54
Taking the LCM of 336,
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1
Taking the LCM of 54,
54 = 2 × 3 × 3 × 3 × 1
Therefore, LCM of 336 and 54 = 3024
HCF of 336 and 54 = 2×3 = 6
Now, the product of 336 and 54 = 336 × 54 = 18144
And the product of LCM and HCF = 3024 × 6 = 18144
Therefore, LCM × HCF = product of the 336 and 54.
问题 3：应用素因数分解法求下列整数的 LCM 和 HCF。
(i) 12、15 和 21
(ii) 17、23 和 29
(iii) 8、9 和 25
解决方案：
(i) 12, 15 and 21
Taking the LCM of 12,
12=2×2×3
Taking the LCM of 15,
15=5×3
Taking the LCM of 21,
21=7×3
Therefore,
HCF of 12, 15 and 21 = 3
LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
Taking the LCM of 17,
17=17×1
Taking the LCM of 23,
23=23×1
Taking the LCM of 29,
29=29×1
Therefore,
HCF of 17, 23 and 29 = 1
LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
Taking the LCM of 8,
8=2×2×2×1
Taking the LCM of 9,
9=3×3×1
Taking the LCM of 25,
25=5×5×1
Therefore,
HCF of 8, 9 and 25 =1
LCM of 8, 9 and 25 = 2×2×2×3×3×5×5 = 1800
问题 4：假设 HCF (306, 657) = 9，求 LCM (306, 657)。
解决方案：
Given: HCF (306, 657) = 9
We know that,
HCF×LCM=Product of the two given numbers
Therefore, by substituting the value we get,
9 × LCM = 306 × 657
LCM = (306×657)/9 = 22338
LCM (306,657) = 22338
Therefore, the LCM (306,657) = 22338
问题 5：检查 6 n是否可以对任何自然数 n 以数字 0 结尾。
解决方案：
If the given number 6n ends with the digit 0, then it should be divisible by 5.
We know that if any number with the unit place as 0 or 5 is divisible by 5.
Therefore,
By prime factorization of 6n = (2×3)n
Since the prime factorization of 6n doesn’t contain prime number 5.
Therefore, 6n cannot end with the digit 0 for any natural number.
问题 6：解释为什么 7 × 11 × 13 + 13 和 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 是合数。
解决方案：
We know by the definition of a composite number, that, the composite number has factors other than 1 and itself.
Therefore, in the given expression
7 × 11 × 13 + 13
By taking 13 as a common factor, we get,
=13(7×11×1+1)
= 13(77+1)
= 13×78 [taking prime factors of 78]
= 13×3×2×13
Therefore, 7 × 11 × 13 + 13 is a composite number.
Now, for the 2nd number
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
By taking 5 as a common factor, we get,
=5(7×6×4×3×2×1+1)
= 5(1008+1)
= 5×1009
Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.
问题 7：运动场周围有一条圆形路径。 Sonia 需要 18 分钟才能打完一轮，而 Ravi 需要 12 分钟。假设它们都从同一点同时开始，并朝同一方向行进。多少分钟后，他们会在起点再次相遇？
解决方案：
Given: Both Sonia and Ravi move in the same direction and at the same time.
Now, the time when they will be meeting again at the starting point can be calculated by finding the LCM of 18 and 12
Therefore,
LCM (18, 12) = 2×3×3×2×1 = 36
Finally, they both will meet again at the starting point after 36 minutes.