问题1.假设数字35a64可被3整除，其中a是一个数字，那么a的可能值是什么?
解决方案：
The number 35a64 is divisible by 3
∵ The sum of its digits will also be divisible by 3
∴ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 (∵ 18 is divisible by 3)
∴ Values of a can be 0, 3, 6, 9
问题2.如果x是一个数字，使得18×71的数字可被3整除，请找到x的可能值?
解决方案：
∵ The number 18×71 is divisible by 3
∴ The sum of its digits will also be divisible by 3
⇒ 1 + 8 + x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27……
∴ x can be 1, 4, 7 which are divisible by 3.
问题3.如果x是66784x的数字，使得它可以被9整除，请问x的可能值是多少?
解决方案：
∵ The number 66784x is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6 + 6 + 7 + 8 + 4 + x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54………
which are divisible by 9
∴ Values of x can be 5 on 9
∴ x = 5
问题4.假设数字67y19可被9整除，其中y是一个数字，那么y的可能值是什么?
解决方案：
∵ The number 67y19 is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6 + 7 + y + 1 + 9 is divisible by 9
⇒ 23 + y is divisible by 9
∴ The numbers greater than 23 are 27, 36, 45,……..
Which are divisible by 9
∴ 23 + y = 27 ⇒ y = 4
问题5.如果3×2是11的倍数，其中x是一个数字，那么x的值是多少?
解决方案：
∵ The number 3×2 is a multiple of 11
∴ It is divisible by 11
∴ The difference of the sum of its alternate digits is zero or multiple of 11
∴ The difference of (2 + 3) and * is zero or multiple of 11
⇒ If x – (2 + 3) = 0 ⇒ x – 5 = 0
Then x = 5
问题6.如果98125×2是一个以x为十进制数的数字，使得它可以被4整除。找到x的所有可能值。
解决方案：
∵ The number 98125×2 is divisible by 4
∴ The number formed by tens digit and units digit will also be divisible by 4
∴ x2 is divisible by 4
∴ The possible number can be 12, 32, 52, 72, 92
∴ Value of x will be 1, 3, 5, 7, 9
问题7.如果x表示数字67×19的数百个位上的数字，使得该数字可以被11整除。找到x的所有可能值。
解决方案：
∵ The number 67×19 is divisible by 11
∴ The difference of the sums its alternate digits will be 0 or divisible by 11
∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11
⇒ 15 + x – 8 = 0, or multiple of 11,
7 + x = 0 ⇒ x = -7, which is not possible
∴ 7 + x = 11, 7 + x = 22 etc.
⇒ x = 11 – 7 = 4, x = 22 – 7
⇒ x = 15 which is not a digit
∴ x = 4
问题8.将981547除以5后，求出余数。不进行实际除法即可。
解决方案：
A number is divisible by 5 if its units digit is 0 or 5
But in number 981547, the units digit is 7
∴ Dividing the number by 5,
Then remainder will be 7 – 5 = 2
问题9.找到51439786除以3后的余数。请执行此操作而不执行实际除法。
解决方案：
In the number 51439786, sum of digits is 5 + 1 + 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.
∴ The sum of digits must be divisible by 3
∴ Dividing 43 by 3, the remainder will be = 1
Hence remainder = 1
问题10：找到余数，当798除以11时不进行实际除法。
解决方案：
Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6
∴ Remainder = 6
问题11.在不进行实际除法的情况下，找到928174653除以11所得的余数。
解决方案：
Let n = 928174653
= A multiple of 11 + (9 + 8 + 7 + 6 + 3) – (2 + 1 + 4 + 5)
= A multiple of 11 + 33 – 12
= A multiple of 11 + 21
= A multiple of 11 + 11 + 10
= A multiple of 11 + 10
∴ Remainder = 10
问题12.给出一个可以被整除的数字的示例：
(i)2但不以4(ii)3但不以6
(iii)4但不等于8(iv)4和8均不等于32
解决方案：
(i)2而不是4
A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.
∴ The number can be 222, 342, etc.
(ii)3而不是6
A number is divisible by 3 if the sum of its digits is divisible by 3
But a number is divisible by 6, if it is divided by 2 and 3 both
∴ The numbers can be 333, 201, etc.
(iii)4而不是8
A number is divisible by 4 if the number formed by the tens digit and one’s digit is divisible by 4 but a number is divisible by 8 if the number formed by hundreds digit, the tens digit, and one digit is divisible by 8.
∴ The number can be 244, 1356, etc.
(iv)4和8，但不是32
A number in which the number formed by the hundreds, tens, and one’s digit, is divisible by 8. It will also divisible by 4 also.
But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400, etc.
问题13.以下哪个陈述是正确的?
(i)如果一个数字可以被3整除，那么它必须可以被9整除。
(ii)如果一个数字可以被9整除，那么它必须可以被3整除。
(iii)如果一个数字可以被4整除，那么它必须可以被8整除。
(iv)如果一个数字可以被8整除，那么它必须可以被4整除。
(v)如果一个数字可以被3和6整除，则可以被18整除。
(vi)如果一个数字可以被9和10整除，则必须被90整除。
(vii)如果数字精确地将两个数字之和除，则必须分别精确地将数字除。
(viii)如果一个数字精确地将三个数字相除，则必须将它们的和精确地相除。
(ix)如果两个数字是共同优先级，则至少其中一个必须是质数。
(x)两个连续的奇数之和始终可被4整除。
解决方案：
(i) False, it is not necessarily that it must divide by 9.
(ii) True.
(iii) False, it is not necessarily that it must divide by 8.
(iv) True.
(v) False, it must be divisible by 9 and 2 both.
(vi) True.
(vii) False, it is not necessarily.
(viii) True.
(ix) False. It is not necessarily.
(x) True.