Find A in the addition, given that B is a natural number.

$\begin{matrix} A + A + A B A \end{matrix}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
5

The sum of three A's is such a number that has A in its ones place.

Therefore, the sum of two A's must be a number whose ones digit is 0.

This happens only for A = 0 and A = 5

If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0.

Given that B is a natural number, therefore B cannot be equal to 0. So we don't consider this possibility.

Hence, A = 5 $\begin{matrix} 5 + 5 + 5 15 \end{matrix}$

Therefore, A= 5 and B= 1

Suggest Corrections

Similar questions

Find A in the addition, given that B is a natural number.

$\begin{matrix} A + A + A B A \end{matrix}$

Find A in the addition, when B=1

$\begin{matrix} A + A + A B A \end{matrix}$

Q. Let

A = [\begin{matrix} s i n θ & 0 0 & - s i n θ \end{matrix}]

. If

A + A^{T}

is a null matrix, then the number of values of

θ

[0, 2 π]

a + b = b + a

is the ______ property under addition of Natural Numbers

Match the following: $\begin{matrix} C o l u m n A & C o l u m n B 1. {1, 2, 4, 8} & A . {x : x i s a n a t u r a l e v e n n u m b e r l e s s t h a n 10} 2. {2} & B . {x : x i s a p r i m e n u m b e r a n d a d i v i s o r o f 8} 3. {2, 4, 6, 8} & C . {x : x i s a d i v i s o r o f 8} 4. {4, 6, 8, 9} & D . {x : x i s a c o m p o s i t e n u m b e r l e s s t h a n 10} \end{matrix}$

Join BYJU'S Learning Program

Find A in the addition, given that B is a natural number. A+A+AB A

Find A in the addition, given that B is a natural number.

$\begin{matrix} A + A + A B A \end{matrix}$