Question

ABC is a 3 digit number such that it's last digit C is divisible by 2;the two digit part BC is divisible by 3;ABC itself is divisible by 4.Tge number of such three numbers is​

Answer 1

Answer:

A

Step-by-step explanation:

You are told that X is a three-digit integer in which each digit is either 1 or 2. There are 8 possibilities for X (8 = 2×2×2), but rather than list out this many possibilities, you might just write something like this: X = abc product of a, b, and c, but rather the three-digit number formed from the digits in that order (a = hundreds, b = tens, c = units). Likewise, you can now write Y = cba.

You need to find the remainder when X is divided by 3. Since divisibility by 3 depends on the sum of the digits, you really just need to find a + b + c, or more precisely, whether this sum is itself a multiple of 3, one more than a multiple of 3, or two more than a multiple of 3.

Statement 1: SUFFICIENT. To express the hundreds digit of XY in terms of a, b, and c, you need to write X and Y in a more formal algebraic way. A three-digit number is 100 times its hundreds digit, plus 10 times its tens digit, plus its units digit. So X = 100a + 10b + c. Likewise, Y = 100c + 10b + a. We now have to multiply these two expressions together. The efficient way to do so is to think of the possible coefficients (10,000; 1,000; 100; 10; and 1) and then match the terms that will create those coefficients.

(100a+ 10b + c)(100c + 10b + a) =10,000ac + 1,000(ab + bc) + 100(a^2 + b^2 + c^2) + 10(ab + bc) + ac.

So you are told that the hundreds digit is 6. Before you match that to the expression above, you have to think about carried digits—could the expressions in the units or the tens place cause a digit to be carried? The answer is no: even if all the digits were 2’s (the maximum), the tens product would only be 8, with no carrying. So we can now say that 6 = a^2 + b^2 + c^2. Since each variable can only be 1 or 2, what are the possible values of the digits? By testing numbers, you can quickly see that exactly one of the digits must be 2; the other two digits must be 1. Thus, the sum of the digits of X is 2 + 1 + 1 = 4, so the remainder after division by 3 is 1.

Statement 2: INSUFFICIENT. Using the same work from Statement 1, and checking that you don’t have to worry about carried digits, you get ab + bc = 4. Factor the left side: b(a + c) = 4. Given the possible digit values of 1 and 2, there are two possible solutions. One solution is b = 2 and a + c = 2, or a = 1 and b = 1. The other solution is b = 1 and a + c = 4, or a = 2 and c = 2. That means that the number is either 121 or 212. If the number is 121, the remainder is 1. If the number is 212, the remainder is 2.

The correct answer is A.

Answer 2

We have a 3 digit number abc that is divisible by 3.

Therefore, we can say that:

(a+b+c) is also divisible by 3 (1)

Now we conclude that, 111 is divisible by 3 and

(a+b+c) is also divisible by 3 (From (1))

Therefore, 111(a+b+c) is divisible by 9.

Hence, abc+bca+cab is divisible by 9.

thanks

hope it helps

:)