Adding Digits Alternately Tests a Number for Eleven

There is. Add the digits in alternate positions, subtract one total from the other, and check whether the answer is divisible by 11.

For $1331$: $(1+3)-(3+1)=0$, which is divisible by 11 — and indeed $1331÷11=121$.

This page covers everything in the CBSE Class 8 Mathematics chapter's second part: the tests for 3 and 9, the test for 11, the algebra that proves them, and cryptarithm puzzles.

Add all the digits. A number is divisible by 3 if its digit sum is divisible by 3, and by 9 if its digit sum is divisible by 9.

Worked example. Test $4287$:

$$4+2+8+7=21$$

Since $21$ is divisible by 3, so is $4287$ — and $4287÷3=1429$. But $21$ is not divisible by 9, so $4287$ is not divisible by 9 either.

Worked example. Test $5832$:

$$5+8+3+2=18$$

Since $18$ is divisible by both 3 and 9, so is $5832$ — and $5832÷9=648$.

Worked example. Test $7415$:

$$7+4+1+5=17$$

Not divisible by 3, so $7415$ is not either.

If the digit sum is still large, repeat the process. For $99999$ the sum is $45$, and $4+5=9$ — divisible by 9.

The relationship between the two tests is worth stating. Every number divisible by 9 is automatically divisible by 3, since 3 is a factor of 9 — but not the reverse, as $4287$ shows. So a number passing the 3 test still needs the 9 test separately.

Take the digits from the right. Add those in the odd positions, add those in the even positions, and subtract one total from the other. If the result is $0$ or a multiple of 11, the number is divisible by 11.

Worked example. Test $1331$. From the right the digits are $1,3,3,1$:

- Odd positions (1st and 3rd): $1+3=4$
- Even positions (2nd and 4th): $3+1=4$

$$4-4=0$$

Divisible by 11, and $1331÷11=121$.

Worked example. Test $61809$. From the right: $9,0,8,1,6$:

- Odd positions: $9+8+6=23$
- Even positions: $0+1=1$

$$23-1=22$$

Since $22$ is a multiple of 11, so is $61809$ — and $61809÷11=5619$.

Worked example. Test $4587$. From the right: $7,8,5,4$:

- Odd positions: $7+5=12$
- Even positions: $8+4=12$

$$12-12=0$$

So $4587$ is divisible by 11, giving $417$.

Worked example that fails. Test $1234$: $(4+2)-(3+1)=6-4=2$. Not a multiple of 11, so $1234$ is not divisible by 11.

A result of 0 counts as divisible, and that is the case most often doubted. Zero is a multiple of 11, since $11\times 0=0$ — so an alternating sum of 0 is a pass, not a failure.

In a cryptarithm, each letter stands for a different digit, and the same letter always means the same digit. Work from the column that gives the most information, usually the units.

Addition puzzle. Find the digit $A$:

$$3A+25=62$$

where $3A$ means the two-digit number with tens digit 3 and units digit $A$.

- Units column: $A+5$ must end in $2$. So $A+5=12$, giving $A=7$ and a carry of 1.
- Tens column: $3+2+1=6$, which matches.

So $A=7$, and checking: $37+25=62$. Correct.

Multiplication puzzle. Find the digit $A$:

$$1A\times 6=96$$

- Units: $A\times 6$ must end in $6$. Testing digits, $A=1$ gives 6 and $A=6$ gives 36 — both end in 6.
- Try $A=1$: $11\times 6=66$, which is not 96.
- Try $A=6$: $16\times 6=96$. Correct.

So $A=6$.

Using a divisibility test in a puzzle. Find the digit $x$ so that $47x$ is divisible by 9. The digit sum is $4+7+x=11+x$, and the next multiple of 9 is 18, so

$$11+x=18\Rightarrow x=7$$

Checking: $477÷9=53$. Correct.

The strategy that works is to start where the choices are fewest and to test the candidates rather than guess. The units column usually narrows a letter to one or two possibilities, and trying each against the rest of the sum settles it — which is exactly what separated $A=6$ from $A=1$ above.