# Divisibility test

Zaza
Mod note: Moved from technical forum section, so missing the usual sections.

Hi am 16yo and i was unable to tackle this quiz even despite trying some online calculators. i hope someone can explain to me step by step. thanks
In each of the following numbers without doing actual division, determine whether the first number is divisible by the second number:

(i) 3409122; 6

(ii) 17218; 6

(iii) 11309634; 8

(iv) 515712; 8

(v) 3501804; 4

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PeroK

Mentor
Mod note: Moved from technical forum section, so missing the usual sections.

Hi am 16yo and i was unable to tackle this quiz even despite trying some online calculators. i hope someone can explain to me step by step. thanks
In each of the following numbers without doing actual division, determine whether the first number is divisible by the second number:

(i) 3409122; 6

(ii) 17218; 6

(iii) 11309634; 8

(iv) 515712; 8

(v) 3501804; 4
A number that is divisible by 2 will have its rightmost digit be even.
A number that is divisible by 4 will have its rightmost two digits be divisible by 4. For example, 1216 is divisible by 4, while 1217 is not.
A number that is divisible by 8 will have its rightmost three digits be divisible by 8. For example,, 124,032 is divisible by 8 (since 032 is divisible by 8), but 124,025 is not.
A number that is divisible by 3 has digits that add to a sum that is divisible by 3. For example the digits of 627 are 6, 2, and 7, which add to 15, which is divisible by 3. This implies that 627 is also divisible by 3.
A number that is even and divisible by 3 is also divisible by 6.

There are other rules, but these should suffice for the problems you posted. Your textbook or class notes should list the rules I showed.

In the future, please limit the number of questions asked in a single post to one or two only.

berkeman
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##1000 = 8 \times 125##, therefore, a number is divisible by ##8## iff its last three digits form a number divisible by ##8##.

##100 = 4 \times 25 \dots##

Mentor
##1000 = 8 \times 125##, therefore, a number is divisible by ##8## iff its last three digits form a number divisible by ##8##.

##100 = 4 \times 25 \dots##
These show the reasoning behind the rules I showed. The rules for divisibility by 3 and by 9 (which I didn't list) involve adding up the digits of the number in question.

Gold Member
You can use rules of divisibility by 2, by 3, then you can conclude divisivility by 6. The two rules are pretty straightforward. You can do similar for 4,8.

Homework Helper
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Divisibility by ##7## is the complicated one.

Zaza
Thank you for helping

Mentor
You can use rules of divisibility by 2, by 3, then you can conclude divisivility by 6. The two rules are pretty straightforward. You can do similar for 4,8.
All of these were discussed in post #2.