A Pattern That Works Four Times Can Still Be Wrong

Easily. The claim *the sum of $k$ consecutive numbers is divisible by $k$* works for $k=1,3,5$ and $7$ — and fails for $k=4$, since $7+8+9+10=34$, which is not divisible by 4.

Only algebra can tell you which cases survive. This page covers everything in the CBSE Class 8 Mathematics chapter's second part: explaining a grid puzzle, the largest product, justifying a divisibility trick, and proving or breaking a pattern.

Name the centre entry with a letter and write every other entry in terms of it. The letters then do the explaining.

The puzzle. In a calendar, draw a box around any $3\times 3$ block of dates. Add all nine. The total is always nine times the centre.

Why. In a calendar each row is $7$ days after the one above, and each column steps by $1$. So with the centre called $n$, the block is

- Top row: $n-8$, $n-7$, $n-6$
- Middle row: $n-1$, $n$, $n+1$
- Bottom row: $n+6$, $n+7$, $n+8$

Adding them, every offset cancels against its opposite:

$$(-8-7-6-1+0+1+6+7+8)=0$$

so the total is simply

$$9n$$

Checking with numbers. Centre $15$ gives the block $7,8,9,14,15,16,21,22,23$:

$$24+45+66=135,\text{and }9\times 15=135$$

The cancelling of offsets is what makes the trick work, and it explains why the rule survives anywhere on the grid. Every entry above the centre is matched by one equally far below it, so their deviations destroy each other — which is also why the same puzzle works on a $5\times 5$ block, giving $25n$.

Put the two largest digits in the tens places, then pair the larger tens digit with the smaller units digit.

Worked example. Use each of $1,2,3,4$ once to form two two-digit numbers with the largest possible product.

The tens places must take $4$ and $3$, since tens are worth ten times as much. That leaves $2$ and $1$ for the units, and there are two ways to finish:

$$41\times 32=1312,42\times 31=1302$$

So **$41\times 32=1312$ is the largest, and the larger tens digit took the smaller unit.

Justifying it algebraically.** Let the tens digits be $a>b$ and the units digits be $p>q$. Compare the two arrangements:

$$(10a+q)(10b+p)-(10a+p)(10b+q)$$

Expanding both products, the $100ab$ terms and the $pq$ terms cancel, leaving

$$10(ap+bq-aq-bp)=10(a-b)(p-q)$$

Since $a>b$ and $p>q$, both brackets are positive, so the difference is positive — meaning $(10a+q)(10b+p)$ is the larger product.

With $a=4$, $b=3$, $p=2$, $q=1$ that is $41\times 32$, exactly as found.

The surprise is that pairing the biggest with the biggest is wrong, and the algebra shows why. Choosing $42\times 31$ puts both large digits in one number, which raises that number but lowers the other more — and the product depends on both.

Write the number in expanded place value form with letters for the digits, then simplify until a factor appears.

Trick 1 — reverse and subtract. Take any two-digit number, reverse its digits and subtract the smaller from the larger. The answer is always divisible by 9.

With digits $a$ and $b$, the number is $10a+b$ and its reverse is $10b+a$:

$$(10a+b)-(10b+a)=9a-9b=9(a-b)$$

A multiple of 9, whatever the digits.

Checking: $72-27=45=9\times 5$, and $a-b=7-2=5$. Correct.

Trick 2 — reverse and add. The sum is always divisible by 11:

$$(10a+b)+(10b+a)=11a+11b=11(a+b)$$

Checking: $72+27=99=11\times 9$, and $a+b=9$. Correct.

Trick 3 — a three-digit number with all digits equal, such as $555$. With digit $a$:

$$100a+10a+a=111a=3\times 37\times a$$

so it is always divisible by 3 and by 37. Checking: $555÷37=15$.

The method is the same each time, and it is worth naming: replace the digits with letters, expand, and factorise. The factor that comes out is the divisor the trick promises — which is why the trick holds for every number of that shape rather than just the ones you happened to test.

Write the general case with letters. If it factorises, the pattern is proved. If it does not, look for the case where the factor is missing.

Proving. The sum of three consecutive integers is divisible by 3.

$$n+(n+1)+(n+2)=3n+3=3(n+1)$$

A multiple of 3 always. Checking: $7+8+9=24=3\times 8$.

Proving again. The sum of five consecutive integers is divisible by 5.

$$n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10=5(n+2)$$

Always a multiple of 5. Checking: $3+4+5+6+7=25$.

Where it fails. The sum of four consecutive integers is divisible by 4.

$$n+(n+1)+(n+2)+(n+3)=4n+6=2(2n+3)$$

Since $2n+3$ is odd, the expression has only one factor of 2 — so it is divisible by 2 but never by 4.

Checking: $7+8+9+10=34$, and $34÷4=8.5$. The claim is false, and $7,8,9,10$ is the counterexample.

The real pattern. The algebra reveals what is actually true: the sum of $k$ consecutive integers is divisible by $k$ when $k$ is odd, and not when $k$ is even. For odd $k$ there is a middle term and the offsets cancel; for even $k$ there is no middle term and a leftover remains.

That is what algebra gives you that examples cannot. Testing $k=1,3,5,7$ would have suggested the claim always held — and only the general expression showed both why it works for odd $k$ and exactly where it breaks.