## You have to use suitable techniques systematically for solving hard math problems quickly

### The problem

How many of the four digit natural numbers divisible by 24 have also 24 appearing in them?

- 26
- 25
- 19
- 15

### Solution to the hard CAT level number system problem

The phrase "24 appearing in them" makes you to promptly break down the possible set of numbers into three groups of the forms,

Group 1: $24xy$

Group 2: $x24y$

Group 3: $xy24$.

You know by this pattern identified that, you have to use the concept of Place Value Mechanism.

**Note:** Effectively you have applied **problem breakdown technique** to break-up the set of possible numbers into *three well-defined groups.*

### Enumeration of possible numbers in Group 1 of form $24xy$

It takes just a few moments to identify the **5 possible numbers** in this group as,

2400, 2424, 2448, 2472, 2496.

**Note:** This is because by *place value breakup*, $24xy=2400+xy$ and $2400$ being divisible by 24, $xy$ must also be divisible by 24.

But **looking ahead** to possible results of enumeration of other two groups for **detecting overlaps**, it is easy to spot 2424 that would be a member of the Group 3 enumeration also.

So at this stage itself you reduce the effective **number of members in the first group to 4.**

By this **overlap analysis and adjustment**, you have taken care that the future enumerations you would do without remembering anything about this adjustment. It would reduce your memory load.

### Enumeration of possible numbers in Group 2 of form $x24y$

On this form you apply problem break-up again by detecting the pattern that,

In $x24y=x000+24y$ by divisibility rule for 8, which is a factor of 24, $24y$ must be divisible by 8. By this condition, only 240 and 248 would be valid and value of

y can only be 0 or 8.

Because of the form of this group you are using the most effective factor breakup of $24=8\times{3}$.

With this reduction of possiblities to the form, $x240$, or $x248$, you would just consider values of $x$ from 1 to 9 and check for each whether $x+y$ is divisible by 3. You could use this stringent condition using the divisibility rule for 3 because,

For $x24y$ to be divisible by 3, the integer sum $x+2+4+y$ must be divisible by 3. As $2+4$ is already divisible by 3,

$x+y$ with y as 0 and 8, must then be divisible by 3.

Enumeration is quick and all in mind resulting in **6 possible combinations** of $x$ and $y$ values,

1 : 18 : valid

2 : 20, 28 : invalid

3 : 30 : valid

4 : 48 : valid

5 : 50, 58 invalid

6 : 60 : valid

7 : 78 : validÂ

8 : 80, 88 ; invalid

9 : 90 : valid.

### Enumeration of possible numbers in Group 3 of form $xy24$

You waste no time to express $xy24=100\times{xy}+24$ using place value break-up.

24 being divisible by 24, $100\times{xy}$ must be divisible by 24.

Identifyng that 100 has the factor 4 in it, this time you use the suitable factor break-up of $24=4\times{6}$ and enumerate just all the possible factors of two digit number $xy$ where $x \neq 0$.

Starting from 12 there can only be 15 such factors up to 96,

12, 18, 24, ...... 96.

Summing up the possible numbers in the three groups, finally total number of 4 digit numbers divisible by 2r4 having also 24 in them is,

$4+6+15=25$.

**Answer:** Option b: 25.

**Concepts and techniques used:** Key pattern identification -- Problem breakdown -- Place value mechanism -- Looking ahead for consequence analysis -- Overlap analysis and adjustment -- Divisibility rules for 8 and 3 -- Using most effective factorization -- Systematic Enumeration -- Systematic problem solving.

This is systematic, confident and quick problem solving using suitable concepts and techniques at each step.