Problems Involving Divisibility Rules: Difficult Problems with Solutions

Solution:
The numbers which are divisible by both 3 and 2 are also divisible by 6. The numbers which end with (0, 2, 4, 6, 8) are divisible by 2, and the numbers for which the sum of all individual digits is divisible by 3 is wholly divisible by 3,
215496 end with "6", hence it is divisible by 2.
And (2+1+5+4+9+6) = 27, which is divisible by 3.
Since 215496 is divisible by both 2 and 3, therefore it is also divisible by 6.

Solution:
5152 is divisible by 2 because the last digit is 2. But (5+1+5+2) = 13, which is NOT divisible by 3.
1980 ends in 0, hence it is divisible by 2. And (1+9+8+0) = 18, which is divisible by 3. Therefore it is divisible by both.
9023 ends in 3, which is not a multiple of 2, hence it is NOT divisible by 2. And (9+0+2+3) = 14, which is NOT divisible by 3.
1980 is the number divisible by both 2 and 3.

Solution:
4+1+4+6 = 15. 15 is divisible by 3 but not divisible by 9.
2+9+5+3 = 19. 19 is neither divisible by 3 nor 9.
1+8+6+3 = 18. 18 is divisible by BOTH 3 and 9. Hence 1863 is divisible by 3 and 9

Solution:
6+3+1+3 = 13. 13 is not divisible by 3.
5+3+1+3 = 12. 12 is divisible by 3. But it ends in 3(not divisible by 2) so it is not divisible by 6.
9+8+7+0 = 24. 24 is divisible by 3. And it ends in 0(divisible by 2). Since it is both divisible by 2 and 3, therefore it is also divisible by 6.

Solution:
1429 ends in the number "9". Therefore it is neither divisible by 5 nor 10.
8625 ends in the number "5". Hence it is divisible by 5 but not 10.
4620 ends in the number "0". Hence it is divisible by both 5 and 10

Solution:
4563 ends in 3 which is not a multiple of 2, hence indivisible by 2. But (4+5+6+3) = 18, which is divisible by 9.
9792 ends in 2, hence it is divisible by 2. And (9+7+9+2) = 27, which is divisible by 9. Therefore it is divisible by both 2 and 9.
2302 ends in 2, hence it is divisible by 2. But (2+3+0+2) = 7, which is indivisible by 9.

Solution:
4460 ends in 0 so it is divisible by 10. But (4+4+6+0) = 14, which is indivisible by 9.
3969 ends in 9, so it is indivisible by 10. But (3+9+6+9) = 27, which is divisible by 9.
1440 ends in 0, so it is divisible by 10. And (1+4+4+0) = 9, which is divisible by 9. Therefore 1440 is divisible by both 9 and 10.

Solution:
5661 ends in 1 so it is NOT divisible by 5. But (5+6+6+1) = 18, which is divisible by 9.
8425 ends in 5 so it is divisible by 5. But (8+4+2+5) = 19, which is NOT divisible by 9.
4590 ends in 0 so it is divisible by 5. And (4+5+9+0) = 18, which is divisible by 9. Therefore 4590 is divisible by both 5 and 9.

Solution:
6723 ends in 3 so it is NOT divisible by 5. But (6+7+2+3) = 18, which is divisible by 3.
4875 ends in 5 so it is divisible by 5. And (4+8+7+5) = 24, which is divisible by 3. Therefore it is divisible by both.
9460 ends in 0 so it is divisible by 5. But (9+4+6+0) = 19, which is NOT divisible by 3.

Solution:
1908 does not end in 0 so it is NOT divisible by 10. 1908 ends in 8(divisible by 2) and (1+9+0+8) = 18 is divisible by 3, hence it is divisible by 6.
5930 ends in 0 so it is divisible by 10. 5930 ends in 0(divisible by 2) but (5+9+3+0) = 17 is NOT divisible by 3, hence it is NOT divisible by 6.
2700 ends in 0 so it is divisible by 10. 2700 ends in 0(divisible by 2) and (2+7+0+0) = 9 is divisible by 3, hence it is divisible by 6. Therefore it is divisible by both.

Solution:
5062 ends in 2 so it is divisible by 2. But NOT divisible by 5.
4390 ends in 0 so it is divisible by 2. And also divisible by 5. Hence it is divisible by both 2 and 5.
7585 ends in 5 so it is NOT divisible by 2. But it is divisible by 5.

Solution:
For a number to be divisible by 6, it must be divisible by both 2 and 3 at the same time.

6843 ends in 3(indivisible by 2). But (6+8+4+3) = 21, which is divisible by 3 but not by 9.
9756 ends in 6 (divisible by 2). And (9+7+5+6) = 27, which is both divisible by 3 and 9. Therefore it is divisible by all 3, 6 and 9.
8433 ends in 3(indivisible by 2). But (8+4+3+3) = 18, which is both divisible by 3 and 9.

Solution:
2952 = 2+9+5+2 = 18. 2952 is divisible by 4 because 52 ÷ 4 = 13. (The last two digits of the number must be divisible by 4 for the whole number to be divisible by 4.)
For a number to be divisible by 12, it must be divisible by both 3 and 4.
2952 is divisible by 3: 2+9+5+2=18, 18 ÷ 3 = 6. We already know that 2952 is divisible by 4 because its last two digits are divisible by 4. Therefore, the correct answer is 2952.

Solution:
Let the number be [tex]\overline{a14b}[/tex], [tex]a\neq 0[/tex].
Divisibility by [tex]5[/tex]: [tex]b=0[/tex] or [tex]b=5[/tex].
[tex]b=0[/tex]: [tex]a+1+4+0=a+5[/tex] must be divisible by [tex]3[/tex], so [tex]a\in\{1, 4, 7\}[/tex].
[tex]b=5[/tex]: [tex]a+1+4+5=a+10[/tex] must be divisible by [tex]3[/tex], so [tex]a\in\{2, 5, 8\}[/tex].
The numbers are [tex]1140[/tex], [tex]4140[/tex], [tex]7140[/tex], [tex]2145[/tex], [tex]5145[/tex], [tex]8145[/tex] – there are [tex]6[/tex] of them.

Solution:
If the numbers are [tex]n[/tex], [tex]n+1[/tex], [tex]n+2[/tex], their sum is [tex]n+(n+1)+(n+2)=3n+3=3(n+1)[/tex].
Since [tex]n[/tex] is odd, [tex]n+1[/tex] is even, so the sum [tex]3(n+1)[/tex] is divisible by [tex]3[/tex] and is even.
The sum is even, so its last digit [tex]a[/tex] is even: [tex]a\in\{0, 2, 4, 6, 8\}[/tex].
Divisibility by [tex]3[/tex]: the digit sum [tex]7+a+9+a=16+2a[/tex] must be divisible by [tex]3[/tex]:
[tex]16+2a\in\{16, 20, 24, 28, 32\}[/tex]
Only [tex]24[/tex] is divisible by [tex]3[/tex], so [tex]a=4[/tex].
Check: [tex]2497+2498+2499=7494[/tex]

Solution:
Three-digit multiples of [tex]9[/tex]: from [tex]108=9\times 12[/tex] to [tex]999=9\times 111[/tex], i.e. [tex]111-12+1=100[/tex] numbers.
A number divisible by both [tex]9[/tex] and [tex]12[/tex] is divisible by their least common multiple [tex]36[/tex]. Three-digit multiples of [tex]36[/tex]: from [tex]108=36\times 3[/tex] to [tex]972=36\times 27[/tex], i.e. [tex]27-3+1=25[/tex] numbers.
Removing them: [tex]100-25=75[/tex].

Solution:
Let the number be [tex]\overline{abc}=100a+10b+c[/tex], and the reversed number [tex]\overline{cba}=100c+10b+a[/tex].
[tex]\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=99(a-c)[/tex]
Since [tex]99=9\times 11[/tex] and [tex]9=3\times 3[/tex], the difference [tex]99(a-c)[/tex] is divisible by [tex]3[/tex], [tex]9[/tex] and [tex]11[/tex].