Quiz Level-2

Hint

Step 1: Multiply the unit digit of the given number by 4.

i.e. 5 × 4 = 20

Step 2: Now add the product obtained in step 1 with the remaining digits of the given number.

i.e., 20 + 435 = 455

Step 3: Repeat step 1 and step 2, until we get the two-digit number.

i.e, 45 + (5 × 4)

⇒ 45 + 20

⇒ 65

Hence, 65 is divisible by 13, and therefore we can conclude that 4355 is divisible by 13.

Hint

As we know, a number is divisible by 12 if it is divisible by both 3 and 4.

Now, we have to check the divisibility rule of 3 and 4 for the given number 10032.

Checking for Divisibility Rule of 3:

10032 = 1 + 0 + 0 + 3 + 2 = 6, which is divisible by 3.

Checking for Divisibility Rule of 4:

The last two digits of 10032 are 32, which is divisible by 4.

Hence, the number 10032 is divisible by both 3 and 4, we can say that the number 10032 is divisible by 12.

Hint

According to the divisibility rule of 15, a numeral is divisible by 15 if it is divisible by both 3 and 5.

Since the unit digit of 1440 is 0, it is divisible by 5.

Also, the sum of digits of 1440 = 1 + 4 + 4 + 0 = 9

Hence, the sum of digits is 9, it is divisible by 3.

Since 1440 is divisible by both 3 and 5, 1440 is divisible by 15.

Hint

The number 2025 is a composite number because it is divisible at least by 3 and 5. See below how many and what are their divisors.

The prime factorization of the number 2025 is written, as the product of powers, as \(3^4 \cdot 5^2\).
The prime factors of 2025 are 3 and 5.
Prime factors do not always represent all divisors of a number. The number 2025 has the following divisors or factors:
\(1,3,5,9,15,25,27,45,75,81,135,225,405\) and 675

These numbers above represent ‘all’ the divisors of 2025 (not only the prime ones). Note that the number 2025 has 14 divisors.

Hint

As we know, the sequence of three-digit numbers that are divisible by 5 is:
100, 105, 110, 115, 120, ……… 995.
Therefore, we can say that the given sequence is in Arithmetic Progression with the first digit being 100 and the common difference being 5.
i.e, a = 100, d = 5 and nth term = 995
Now, we need to find the number of three-digit numbers that are divisible by 5.
Therefore,
995 = 100 + (n-1)5
995 = 100 + 5n – 5
995 = 95 + 5n
5n = 995 – 95
5n = 900
Therefore, n = 900/5 = 180.
Therefore, the number of 3-digit numbers that are divisible by 5 is 180.

Arithmetic Progression:
An arithmetic progression is a sequence of numbers such that the difference \(d\) between each consecutive term is a constant.
\(a, a+d, a+2 d, a+3 d, \ldots \ldots\)
The \(n^{\text {th }}\) term, \(a_n=a+(n-1) d\)
Sum of first \(n\) terms, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)
\(
=\frac{n}{2}[a+1]
\)

Hint

According to the rule, if a number is divisible by both 2 and 9, then it is divisible by 18.
Because 2464 is an even number, it is divisible by 2.
Sum of the digits 2464 :
2 + 4 + 6+ 4 = 16
Because 16 is not a multiple of 9, 1458 is not divisible by 9.
So, 1458 is not divisible by 18.

Hint

According to the rule, if the last two digits of a number are zeroes or the number formed by the last two digits is a multiple of 25, then the number is divisible by 25.
In 4500, the last two digits are zeroes.
So, 4500 is divisible by 25.

Hint

Step-by-step explanation:
\(
\begin{aligned}
& x_1=84 \\
& a_1=84 / 6=14 \\
& b_1=84 / 7=12
\end{aligned}
\)

Hint

Any number ending in 1 , when raised to any power, still ends in 1. Can you see why? So both \(121^{13}\) and \(101^4\) end in 1 . This means that their difference ends in 0 , which is divisible by 2 . So \(121^{13}-101^4\) is divisible by 2.

Hint

In algebra, we learn about factoring the difference of two squares, \(x^2-y^2=(x-y)(x+y)\). Using this formula here gives \(326^2-\) \(325^2=(326-325)(326+325)=(1)(651)=651\). This is divisible by 3 because \(6+5+1=12\).