Given that $a$ and $b$ are digits such that $\overline{120ab}$ is a multiple of 3, 5 and 7. What is the value of $\overline{ab}$ ?

**
Details and assumptions
**
:

$\overline{abc}$ means $100a + 10b + 1c$ , as opposed to $a \times b \times c$ . As an explicit example, for $a=2, b=3, c=4$ , $\overline{abc} = 234$ and not $2 \times 3 \times 4 = 24$ .

The answer is 75.

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Great! Using all 3 numbers together makes it much more direct to approach the problem.

As a side note, there is at most one number in the range $100A$ to $100A + 99$ which is a multiple of 105.

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Is there any easy solution other than the Hit and Miss technique

Nehal Arifen
- 5 years ago

In order for $\overline{120ab}$ to be divisible by 5, either $b=0$ or $b=5$ . In order for the number to be divisible by 3, the sum of the digits must be divisible by 3, which means that $a+b$ must be divisible by 3. We can split this problem into two cases; if the units digit $b$ is 5 or 0.

**
Case 1:
**
$b=0$
If
$b=0$
, then
$a$
must be divisible by 3 in order for the number
$\overline{120ab}$
to be divisible by 3. Why? we need the sum of the digits to be divisible by 3, and since
$1+2+0+0$
is divisible by 3, we need
$a$
to be divisible by 3 as well. The digits that are divisible by 3 are 0, 3, 6, and 9. We can test each of these numbers
$\overline{120ab}$
where
$a=0, a=3, a=6, a=9$
and
$b=0$
to see whether they are divisible by 7.

$\frac{12000}{7}$ is not an integer.

$\frac{12030}{7}$ is not an integer.

$\frac{12060}{7}$ is not an integer.

$\frac{12090}{7}$ is not an integer.

Therefore, none of these cases work, so $b$ cannot be 0.

**
Case 2:
**
$b=5$
If
$b=5$
, then in order for the number to be divisible by 3,
$a$
must be equal to 1 (mod 3). The digits that are 1 mod 3 are 1, 4, and 7. We can again test them individually.

$\frac{12015}{7}$ is not an integer.

$\frac{12045}{7}$ is not an integer.

$\frac{12075}{7}=1725$ , which is an integer! Our answer is, therfore, $\boxed{75}$

**
Alternative method:
**
Since 12000 is equal to 2 (mod 7), we need
$\overline{ab}$
to be equal to 5 (mod 7) in order for
$12000+\overline{ab}=\overline{120ab}$
to be divisible by 7.

The numbers less than 100 that are equal to 5 mod 7 are:

$05, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, \text{ and } 96$ .

However, since we want our number to be divisible by 5, we can eliminate most of these choices: the last digit must be equal to 0 or 5. Now, the numbers that work are

$05, 40, \text{ and } 75$ .

Now, we need our number to be divisible by 3. Remember that if the sum of the digits of a number is divisible by 3, then the number is as well. Out of these three numbers, 75 is the only one that is divisible by 3.

Checking our answer, we see that $12075$ is divisible by 3, 5, and 7, so our answer is equal to $75$ .

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On a sidenote, I accidentally clicked back or something, causing my whole solution to disapprear...I had to write it again :P

If I am not very clear on things, tell me; I will try my best to explain. (I lost my temper a little bit when everything disappeared, so I might not have been as thorough as the first time)

William Cui
- 7 years, 7 months ago

$\overline {120ab} \pmod {5} \equiv b \pmod {5} \Rightarrow b\pmod {5} \equiv 0$

hence, $b=0,5$

case 1: $b=0$

$\overline {120ab} \pmod {3} \equiv a+b \pmod {3} \Rightarrow a+b \pmod {3} \equiv 0...(i)$

hence, $a \pmod{3} \equiv {0}$ $(b=0)$

a=0,3,6,9

$\overline {120ab} \pmod {7} \equiv 3a+b+2 \pmod {7} \Rightarrow 3a+b+2 \pmod {7} \equiv 0...(ii)$

but,none of the values of $a$ satisfy $3a+2 \pmod {7}=0$ $(since,b=0)$

hence, $b=5$

case 2: $b=5$

from equation (i), $a+5 \pmod {3} \equiv 0 \Rightarrow a \pmod {3} \equiv 1$

hence, $a=1,4,7......(iii)$

from equation (ii) $3a \pmod {7} \equiv 0$ $(b=5)$

or, $a=0,7....(iv)$

from $(iii)$ and $iv)$ we see that $a=7$

hence, $a=7,b=5$

$\overline {ab}= \fbox{75}$

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CAN YOU PLEASE SHOW THE SOLUTION?

unnati bindal
- 7 years, 5 months ago

can you explain how 75 has come?

sujit kumar
- 7 years, 2 months ago

But when
**
b
**
=
$5$
or
$0$
only then it is divisible by
$5$

We know the divisibility rule of
$3$
.The sum of the digits is divisible by
$3$
.

So,when
$b=5$
there are
$3$
choices for
**
a
**
.=
$3$
or
$6$
or
$9$
so The possibilities are are:
$12035,12065,12095$

when
$b=0$
there are
$3$
choices for
**
a
**
.=
$1$
or
$4$
or
$7$
so The possibilities numbers are:
$12015,12045,12075$

But the number must be divisible by
$7$
.

but among those
$6$
numbers there are only one number which is divisible by
$3,5,7$

It is
$\boxed{12075}$

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*
105+30+ab
so,
dividing 114
*
105+30+ab by 105(lcm),we get;
114+(30+ab)/105.
so remainder =0 for complete divisibility,
hence 30+ab/105=1
ab=105-30=75

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amazing!!!

John Gray
- 7 years, 7 months ago

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We have the following equations, using divisibility tests by 3, 5, 7: $b\equiv 0 (mod 5),a+b\equiv 0 (mod 3), 3a+b+2\equiv 0 (mod 7)$ .

Then there exists an integer $k$ such that $b=5k$ .

Then $a+5k\equiv 0 (\mod 3)$ and this is $a+2k\equiv 0(\mod 3)$ thus $a\equiv k (\mod 3)$ .

We know that $b=0$ or $b=5$ and then $k=0$ or $k=1$ .

If $k=0$ then $b=0$ and:

$3a+b+2\equiv 0 (\mod 7)$ becomes $3a\equiv 5 (\mod 7)$ .

The congruence $3a\equiv 5 (\mod 7)$ implies $a\equiv 4 (\mod 7)$ , so $a=4$ .

As $a\equiv k (\mod 3)$ , we have $a=3$ or $a=6$ or $a=9$ (Or 0), so this is a contradiction.

Then $k=1$ . And so we have $b=5$ .

The congruence $3a+b+2\equiv 0 (\mod 7)$ becomes $3a\equiv 0 (\mod 7)$ so $a\equiv 0 (\mod 7)$ and $a=7$ . But also $a\equiv k (\mod 3)$ so $a\neq 0$ .

So $a=7,b=5$ , $\overline{ab}=75$ .

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First we note that since a and b are DIGITS, the greatest possible value of the number $\overline{120ab}$ is $12099$ . Also note that the LCM of $3, 5, 7$ is $105$ . Dividing and taking the floor:

$\lfloor \dfrac{12099}{105} \rfloor = 115$ . Now multiplying by the LCM, we get:

$115\cdot 105 = 12075 \implies \boxed{75}$ , and we are done.

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I know its wrong...

Shivanshu Siyanwal
- 7 years, 7 months ago

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because 120ab is a multiple of 3, 5 and 7 ------> 120ab is a multiple of 105

120ab = (floor (12000/105) + 1) x 105

```
=( floor ( 114,28) + 1) x 105
= 115 x 105
= 12075
```

then, we get ab = 75

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If a number is divisible by 5, then the last digit will be either 0 or 5. b is the last digit and is either 0 or 5. If a number is divisible by 3, then the sum of the digits of the number should be a multiple of 3. If b is 0, then 1+2+0+a+0=multiple of 3. Therefore 'a' can be 0, 3, 6 or 9. The number could be 12030, 12060, 12060 or 12090. If b is 5, then 1+2+0+a+5=multiple of 3. Therefore 'a' can be 1, 4 or 7. The number could be 12000, 12030, 12060, 12090, 12015, 12045, 12075 Now we could try dividing each number by 7 to see which one is divisible by 7. 12075 is the only number that divides without a remainder and hence its last 2 digits 75 are the answer to this question.

Easier way for smarter people:
If a number is divisible by 3, 5 or 7, it should be divisible by 3
*
5
*
7=105. We can guess about how many times we need to multiply it to get something between 12000 and 12100. We use guessing after that to figure out the answer.

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For a number to be a multiple of 3, 5, and 7, it must be a multiple of 105, or $3 \times 5 \times 7$ . As there are only 100 possibilities for what $120ab$ is, there must be only one of those that can be a multiple of 105.

Let's divide 12000 and 12099 by 105 and see if we can get a whole number:

$12000÷105≈114.3$

$12099÷105≈115.2$

Now we know that the number divided by 105 has to be 115.

So the number is $115 \times 105 = 12075$ .

$ab$ is $\boxed{75}$ .