Test: Number bases - 2 - JAMB MCQ

Given, 5^{(x + 3)} = 25^{(3x - 4)}

We can write it as -

5^{(x + 3)} = 5^{2 x (3x - 4)}

Or,

x + 3 = 2(3x - 4)

x + 3 = 6x - 8

or, 5x = 11

So, x = 11/5

According to the question when 'n' is divided by 5, we will get the remainder 2

n/5 = remainder 2

On putting n =7, and on dividing it with 5 we will get remainder 2

So, n = 7, and n² = 49

n²/5 = 49/5 = remainder 4

A number is divisible by 11 if the difference of the sum of digits on odd places and the sum of digits on even places is zero or divisible by 11. So, we have the number A381.

Hence, (A + 8) - (3 + 1) = either 0 or multiple of 11

If we put A = 7, then we get

(7 + 8) - (3 + 1) = 15 - 4 = 11 (Which is divisible by 11).

Suppose the numbers are P and Q

So, according to the question -

P + Q = a

P * Q = b

Sum of reciprocals of P and Q is = 1/P + 1/Q

= Q + P/PQ

= a/b

Let's examine each option individually,

√0.0004 = 0.02

√0.0121 = 0.11

(0.1)² = 0.1

0.12 = 0.12

From the above examination, we can see 0.12 is the largest among others.

We can write 6³⁶/215 as

(6³)¹²/215

Or, we can say 216¹²/215

We will always get remainder 1 on dividing 216 by 215

= 1¹²/215

So, the remainder will be 1.

The largest four-digit number is = 9999, and on dividing it with 66, we will get 33 as the remainder.

So, the largest 4-digit number divided by 66 = 9999 - 33 = 9966

Let the number be X.

According to the question,

2/3 * 3/4 * X = 34

6/12 * X = 34

Or,

1/2 * X = 34

So, X = 68

Now, 20% of 68 is = 68 * 20/100 = 13.6

Let the number be X.

So according to the question -

7X - 3X = 36

4X = 36

Or,

X = 36/4 = 9

So, the required number is 9.

A number is divisible by 9 if the sum of its digits is divisible by 9, e.g., 117. So, we have the number 381A.

Hence, 3 + 8 + 1 + A = multiple of 9

If we put A = 6, then we get

3 + 8 + 1 + 6 = 18 (which is a multiple of 9)

To compare fractions, we can first find a common denominator for all the fractions. The least common multiple of 33, 13, and 15 is 429.

Now, we can convert each fraction to an equivalent fraction with a denominator of 429:

1. 14/33 = (14*13)/(33*13) = 182/429
2. 7/13 = (7*33)/(13*33) = 231/429
3. 11/13 = (11*33)/(13*33) = 363/429
4. 8/15 = (8*13*3)/(15*13*3) = 312/429

Comparing the numerators, we can see that 182 is the smallest, so the smallest fraction among the given options is 14/33 (option 1).

We have 2153¹⁶⁷, and the unit digit is = 3¹⁶⁷

As we know,

3¹ = 3 = unit digit is 3

3² = 9 = unit digit is 9

3³ = 27 = unit digit is 7

3⁴ = 81 = unit digit is 1

The cycle will continue. On the dividing the power of 3 by 4, we will get -

167/4 = remainder is 3

So, 3³ gives us the unit digit 7. So, the unit digit of 2153¹⁶⁷ is 7.

First we need to find the least number, so we have to find out the LCM of 8, 12, 16, and 20.

8 = 2 x 2 x 2

12 = 2 x 2 x 3

16 = 2 x 2 x 2 x 2

20 = 2 x 2 x 5

LCM = 2 x 2 x 2 x 2 x 3 x 5 = 240

240 is the least number that is exactly divisible by 8, 12, 16, and 20.

So, the required number that will leave remainder 5 is -

240 + 5 = 245

A number, which is divisible by 3, 5, and 9, is also divisible by 45.

A number is divisible by 3 if the sum of its entire digits is divisible by 3.
A number is divisible by 9 if the sum of its entire digits is divisible by 9.
The number which ends with 0 or 5 is divisible by 5.

The number 306990 fulfils all the requirements, so the answer is 306990.

The given number is divisible by 6, so it would be divisible by 2 and 3. As the last digit is 2, whatever be the value of X, it would be divisible by 2.

Now, 7 + X + 2 = 9 + X, must be divisible by 3.

So, X = 3 makes the number divisible by 3, so 3 is the required digit.