Question

Find the value of the letter ab×3=cab

Answer 1

Answer:

answer is c=3

Explanation:

ab×3=cab

3=cab/an

3=c

Answer 2

Answer - First let us list which all numbers when multiplied by 3 have the same number in units place.

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=0

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 5

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5Now A has to be a number such that

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5Now A has to be a number such thatA × 3 + 1 = CA

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5Now A has to be a number such thatA × 3 + 1 = CABut no such A exist where adding 1 to the product if the number with 3 gives the number itself in the ones place.

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5Now A has to be a number such thatA × 3 + 1 = CABut no such A exist where adding 1 to the product if the number with 3 gives the number itself in the ones place.So B ≠≠5

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5Now A has to be a number such thatA × 3 + 1 = CABut no such A exist where adding 1 to the product if the number with 3 gives the number itself in the ones place.So B ≠≠5Therefore, AB = 50

First let us list which all numbers when multiplied by 3 have the same number in units place.3×0=03×5=15So B can be either 0 or 5Let B = 0Now A also has to be one of them. But A can't be 0 so A has to be 550 × 3 = 150Let B = 5Now A has to be a number such thatA × 3 + 1 = CABut no such A exist where adding 1 to the product if the number with 3 gives the number itself in the ones place.So B ≠≠5Therefore, AB = 50And CAB = 150

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