1, 2, 0, a, b

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 $\boxed{75}$ .

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.

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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.

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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, $\boxed{75}$

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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 $75$ .

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

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, $a=7,b=5$

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

ab = 75

If the number is multiple of $3$ it must be divisible by $3$ .Again If the number is multiple of $5$ it must be divisible by $5$ .
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 $\boxed{12075}$

take lcm of 3,5,7=105 so , 120ab can be written as 12000+ab=114 105+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

If the number is divisible by 5, then the last digit must be either 0 or 5. The divisibility rule of 7 states that if you double the last digit and subtract the product from the rest of the number and the resulting difference is divisible by 7, then the whole number is divisible by 7. If we double 0, we get 0 so if we have b=0, then $\overline{120a}$ must be divisible by 7. Because 1+2+0=3 (multiple of 3), then a must be a multiple of 3 if b=0. There is no a that satisfies that condition. This means that b must equal 5. Following the same logic as above, we get $\overline{ab}=\boxed{75}$

b must either be 5 or 0 since it is a multiple of 5. a + b must be a multiple of 3 since ab is a multiple of 3. 2000 - ab = multiple of 7. since 14000 - 120ab = 14000 - 12000 - ab This multiple of 7 can either be 1995, 1960, 1925 since they are multiples of 7 and 5. If we take ab for each value, we will get 5, 40, 75 respectively. And the only answer that matches multiple of 3 and 5 is actually 75.

To solve this question, we take the number with easier multipliers:5, which is 12005, 12010, 12015, ..., 12095. Then the multiplier of 3 is known by adding all the digits of the numbers and see if it is divisible by 3. From the numbers 12015 to 12095 which is taken from the number 5 multipliers, only 12015, 12030, 12045, 12060, 12075 and 12090 follows the rule of multiplier 3 and 5, or simply the multiplier of 15. Let's say 12015 => 1+2+0+1+5=9, which is divisible by 3. Then from the 6 choices, divide them by 7 and pick the one with no decimal from the division of 7, which is 12075. So the answer should be 75.

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

ab=75.

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.

In order for a number to be a multiple of 5, it must end with 5 or 0. In this case, it ends with 5. Thus, b is 5. Since the digits in a number divisible by 3 must add up to be a multiple of 3, you can try out the different possibilities: (1+2+0+1+5=9; 1+2+0+4+5=12; 1+2+0+7+5=15) 12015/7=1716 3/7; 12045/7=1720 5/7; and only 12075/7=1725. Hence, ab is 75.

120ab is a multiple of both 3,5 and 7. Finding the value of ab so, we take lowest common factor of 3,5 and 7 = 75 thus, the value of ab is 75 because 120ab can only be divided by 3,5 and 7 unless they have a lowest common factor in it.

For it to be a multiple of 3,5 and 7 the sum of the numbers should be a multiple of 3 and the last number should be 0 or 5.So there are two possiblities of b i.e; 0 and 5.If b=0 then the possible values of a are 0,3,6,9.If b=5 possible values of a are 1,4,7.Substituting these values we see that only the number 12075 is a multiple of these three numbers.So 12075

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

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.