Is the units and tens digit of a two digit number are Y and X respectively then the number will be?

Here, we will see how to find the last two digits of a large number. In the previous post we tried to understand calculating the units digits of large numbers like $2014^{2012}$, $1453^{71}$ etc. In this post,

  • We will first understand how to find the last two digits of a product.
  • And then we will try to find the last two digits ( i.e digits in tens and units place) of a large number in the form ${x^y}$.

At the end of this post, take the quiz to test your understanding of the techniques to find the last two digits of large numbers in the form $x^y$

Video :

The video contains on explanation on how to find the last two digits of a number raised to power based on the units digit of the number.

Understanding the basics – Last two digits of a product

Last two digits of a number is basically the tens place and units place digit of that number.  So given a number say 1439, the last two digits of this number are 3 and 9, which is pretty straight forward. Now, how do we find the last two digits in the product of 1439 x 2786? One feasible approach is using the vertical and cross wise technique of multiplication.

In the product of two numbers say A and B, (in our case A is 1439 and B is 2786). If a and b, respectively represent the digits in the ten’s place and one’s place of A and similarly c and d respectively represent the digits in the ten’s place and one’s place of B, then

  1. Units digit of A x B is given by the units digit in the product of b and d. If the product of b and d results in more than 1 digit, the excess digit will be carried over to the left. i.e in 1439 x 2786, multiply 9 and 6 which gives 54. Here 4 forms the units digit of 1439 x 2786 and 5 goes as carry to Step 2.
  2. In this step, we cross multiply a and d & c and b, and then add the resulting products. i.e in 1439 x 2786 => 3×6 + 9×8 = 90
  3. If a carry is generated in Step 1, add that with the result obtained in Step 2. i.e 5 + 90 = 95
  4. The units digit in the result obtained in Step 3 forms the tens digit in the product of A and B. i.e the units digit in 95 which is 5 becomes the tens digit of 1439 x 2786.

Hence the last two digits of 1439 x 2786 is 5 and 4.

How to find the Last Two Digits of Number raised to Power

Let the number be in the form ${x^y}$. Based on the value of units digit in the base i.e x, we have four cases

Case 1: Units digit in x is 1

If x ends in 1, then x raised to y, ends in 1 and its tens digit is obtained by multiplying the tens digit in x with the units digit in y.

Example 1: Find the last two digits of ${91^{246}}$

Since the base 91 ends in 1, ${ 91^{246}}$ ends in 1 and the tens place digit is obtained from the units digit in 9×6 which is 4. Hence the last two digits of ${ 91^{246}}$ are 4 and 1.

Case 2: Units digit in x is 3, 7 or 9

In this case we will convert the base so that it ends in 1, after which we can use Case 1 to calculate units and tens place digits. i.e

When x ends in 9 ${(..9)^{y}}$

Raise the base by 2 and divide the exponent by 2 => ${(..9^2)^{y/2}}$

Number ending in 9 raised to 2 ends in 1 => ${(..1)^{y/2}}$

Since the base now ends in 1, Tens digit and Units digit is calculated using the steps in Case 1.

Example 2:  Find the last two digits of ${(79)^{142}}$  => ${(79^2)^{142/2}}$ => ${(..41)^{71}}$

Now, units digit of a number ending in 41 to the power of 71 is 1 and its tens digit is obtained by multiplying 4 and 1 which is 4.

Hence, the last two digits of ${(79)^{142}}$ are 4 and 1.

Example 3: Find the last two digits of ${(79)^{143}}$  => $(79)^{142}$ x ${(79)^1}$

$(79)^{142}$ ends in 41 (From previous example) and ${(79)^1}$ ends in 79. Hence, the product of numbers ending in 41 and 79 ends in 39, which implies the last two digits of ${(79)^{143}}$ are 3 and 9.

When x ends in 3 ${(..3)^{y}}$

Raise the base by 4 and divide the exponent by 4 => ${(..3^4)^{y/4}}$

Number ending in 3 raised to 4 ends in 1 => ${(..1)^{y/4}}$

Since the base now ends in 1, Tens digit and Units digit is calculated using the steps in Case 1.

Example 4: Find the last two digits of ${(43)^{76}}$  => ${(43^4)^{76/4}}$ => ${(..01)^{19}}$

Now, units digit of a number ending in 01 to the power of 19 is 1 and its tens digit is obtained by multiplying 0 and 9 which is 0.

Hence, the last two digits of ${(43)^{76}}$ are 0 and 1.

When x ends in 7 ${(..7)^{y}}$

Raise the base by 4 and divide the exponent by 4 => ${(..7^4)^{y/4}}$

Number ending in 7 raised to 4 ends in 1 => ${(..1)^{y/4}}$

Since the base now ends in 1, Tens digit and Units digit is calculated using the steps in Case 1.

Example 5: Find the last two digits of ${(17)^{256}}$  => ${(17^4)^{256/4}}$ => ${(..21)^{44}}$

Units digit of a number ending in 21 to the power of 44 is 1 and its tens digit is obtained by multiplying 2 and 4 which is 8.

Hence, the last two digits of ${(17)^{256}}$ are 8 and 1.

Case 3: Units digit in x is 2, 4, 6 or 8

If x ends in 2, 4, 6, 0r 8, we can find the last two digits of the number raised to power with the help of the following points :

  • ${(2)^{10}}$ ends in 24
  • ${(24)^{odd\ number}}$ ends in 24
  • ${(24)^{even\ number}}$ ends in 76
  • ${(76)^{number}}$ ends in 76

Example 6: Find the last two digits of 2 raised to 1056″

${(2)^{1056}}$ => $(2^{10})^{105}$ x ${(2)^{6}}$

Here, 2 raised to 10 ends in 24 and 24 raised 105, which is an odd number, ends in 24. Also 2 raised to 6 ends in 64. Using the vertical and cross-wise multiplication technique last two digits of the product of the numbers ending in 24 and 64 are 3 and 6.

Case 4: Units digit in x is 5
  • The digit in the tens place is odd and the exponent y is odd, then the number ends in 75.
  • If the digit in the tens place is odd and the exponent y is even, then the number ends in 25.
  • Or if the digit in the tens place is even and the exponent y is odd, then the number ends in 25.
  • If the digit in the tens place is even and the exponent y is even, then the number ends in 25 .

Hence when the exponent and the digit in the tens place of the base are odd, the number raised to power ends 75, in other cases it ends in 25.

Example 7: Find the last two digits of ${(65)^{243}}$

Since the digit in the tens place of the base is even and the exponent is odd, last two digits of ${(65)^{243}}$  are 2 and 5.

Example 8: Find the last two digits of ${(135)^{1091}}$

Since the digit in the tens place of the base is odd and the exponent is odd, ${(65)^{243}}$  ends in 75.

Quiz : Test your Understanding

What Next?

Can you calculate the highest power of  15 in 24! or number of zeros at the end of 1000!? Click on the post below to learn more about highest power of numbers in a factorial.

Is the units and tens digit of a two digit number are Y and X respectively then the number will be?

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

If $x$ and $y$ are the tens and the units digit, respectively, of the product $725{,}278×67{,}066$, what is the value of $x+y$ ?

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 9 of Section 4 of Practice Test 1. Those questions testing our knowledge about Integers can be kind of tricky, but never fear, PrepScholar has got your back!

Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.

We want to multiply two numbers together and look at what numbers are in the units and tens places, which will involve things we’ve learned about Integers.

What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

  1. We will multiply a $6$-digit number by a $5$-digit number
  2. We want to know that product’s sum of its units and tens digits

Develop a Plan

Let’s attack this problem with a top-down approach. We want to know the sum of the units and tens digits of the product of the numbers $725{,}278·67{,}066$. There is one solution readily available to us: we could just multiply the numbers together to get the units and tens digits of the product. That does seem like a time-consuming process, especially if our calculator can’t handle such a large product (like, say, the measly GRE calculator provided to us on test day). So let’s try to find another more clever approach to solving this question.

We might remember from when we first learned long multiplication that the units digit of a product is only determined by the units digits of the numbers being multiplied. Also, the tens digit of a product is determined by the units and tens digits of the numbers being multiplied. If we’re curious about why this is, then let’s feel free to look at it in more depth. Otherwise we can just go ahead and solve this question.

Concept Refresher – Units and Tens Digits for Multiplication

Since we only care about the units and tens digits, perhaps we would be better off looking at numbers specifically based upon their digits, so let’s split one of the numbers up by its digits. We know that each digit represents a factor of $10$, so let’s work on splitting up one of the numbers by writing it as the sum of factors of $10$. It doesn’t matter which number we do this for, so we’ll arbitrarily select $725{,}278$. So let’s first write it as the sum of factors of $10$:

$$725{,}278 = 8 + 70 + 200 + 5{,}000 + 20{,}000 + 700{,}000$$

Now let’s write out the product that we’re looking for using this:

$$725{,}278·67{,}066 = 67{,}066·(8 + 70 + 200 + 5{,}000 + 20{,}000 + 700{,}000)$$

And we know that we can use distribution to rewrite this as the sum of multiple products, by distributing the $67{,}066$ inside the parentheses.

$$725{,}278·67{,}066 = 67{,}066·8 + 67{,}066·70 + 67{,}066·200 + …$$

We know that when we multiply two numbers together, any $0$’s at the end of the numbers will be there in the product. So for example, $5·700 = 5·7·100 = 3{,}500$. So for the three products that involve $200$, $5{,}000$, $20{,}000$ and $700{,}000$, we know that they will all have $0$’s for their units and tens digits. This would also hold true for later when we multiply by $67{,}066$, as we could also split this number up by its digits, focusing on the multiplication with the units and tens digits. So they will add nothing to the sum of the units and tens digits for $725{,}278·67{,}066$. For these reasons, we’ll split both numbers up keeping the units and tens digits separate from the rest of the number, then multiply them together. Now let’s get back to the question at hand.

Solve the Question

So we know that we only need to use the tens and units digits in this multiplication to get the tens and units digits of the product, so let’s select the tens and units digits of the two numbers and multiply them together.

$$78·66 = 5{,}148$$

So the units and tens digits are $4$ and $8$. Since $4+8=12$, the sum of the units and tens digits is $12$, and the correct answer is A, $12$.

What Did We Learn

An important lesson to learn from this problem is that we did NOT have to do the entire product just to find the sum of the units and tens digits. So now we know that for future products where we only want to know the units and tens digits, we can just ignore anything beyond the tens digit.

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