I would recommend you use the Bridge or the Vitruvian man method for 3 digit multipliers to reduce the cognitive load.

UT Method is recommended for two digit multipliers. However it is still possible to do the UT method with the a 3 digit multiplier.

Here is the illustration of the Method

If you want to multiply 6671 × 386 here are the steps.

Step 1: Add the same number of zeros as the number of digits in the multiplier.

So the multiplier 386 has 3 digits. So we add three zeroes in front of 6671 to get 000671

0 0 0 6 6 7 1 x 386

Step 2: Sum of two pair products is a single digit of the answer.

Use your middle and forefinger to keep track of whether to calculate U or the T. Place each finger in the middle of two pairs.

Step 2.1

Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
Using 8 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U8T8 = 00 + 00).
Using 3 in 386, the pair product of 0 and 6 is 0 + 1 = 1 (U3T3 = 00 + 18).
Adding the three number together we get 0 + 0 + 1 = 1.

U6 T6
=   U8 T8
=          U3 T3
0    0     0    6     6     7    1 x 386 = 1 _ _ _ _ _ _

Step 2.2

Moving to the next set of digits

Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
Using 8 in 386, the pair product of 0 and 6 is 0 + 4 = 4 (U8T8 = 00 + 48).
Using 3 in 386, the pair product of 6 and 6 is 8 + 1 = 9 (U3T3 = 18 + 18).
Adding the three number together we get 0 + 4 + 9 = 13.

We attach 3 in 13 as the next digit. But 13 has two digits so we carry over the 1 in 13. So the last digit from the previous step which is 1 becomes 2 (got by carrying over 1+1).

=   U6 T6
=         U8 T8
=                 U3 T3
0    0     0    6     6     7    1 x 386 = 2 3 _ _ _ _ _

Step 2.3

Moving to the next set of digits

Using 6 in 386, the pair product of 0 and 6 is 0 + 3 = 3 (U6T6 = 00 + 36).
Using 8 in 386, the pair product of 6 and 6 is 8 + 4 = 12 (U8T8 = 48 + 48).
Using 3 in 386, the pair product of 6 and 7 is 8 + 2 = 10 (U3T3 = 18 + 21).
Adding the three number together we get 3 + 12 +10 = 25.

We attach 5 in 25 as the next digit. But 25 has two digits, so we carry over the 2 in 25. So the last digit from the previous step which is 3 becomes 5 (got by carrying over 3 + 2).

=          U6 T6
=                 U8 T8
=                        U3 T3
0    0     0    6     6     7    1 x 386 = 2 5 5 _ _ _ _

Step 2.4

Moving to the next set of digits

Using 6 in 386, the pair product of 6 and 6 is 6 + 3 = 9 (U6T6 = 36 + 36).
Using 8 in 386, the pair product of 6 and 7 is 8 + 5 = 13 (U8T8 = 48 + 56).
Using 3 in 386, the pair product of 7 and 1 is 1 + 0 = 1 (U3T3 = 21 + 03).
Adding the three number together we get 9 + 13 +1 = 23.

We attach 3 in 23 as the next digit. But 23 has two digits, so we carry over the 2 in 23. So the last digit from the previous step which is 5 becomes 7 (got by carrying over 5 + 2).

=                 U6 T6
=                       U8 T8
=                              U3 T3
0    0     0    6     6     7    1 x 386 = 2 5 7 3 _ _ _

Step 2.5

Moving to the next set of digits

Using 6 in 386, the pair product of 6 and 7 is 6 + 4 = 10 (U6T6 = 36 + 42).
Using 8 in 386, the pair product of 7 and 1 is 6 + 0 = 6 (U8T8 = 56 + 08).
Using 3 in 386, the pair product of 1 and _ is 3 + _ = 3 (U3T3 = 03 + –-).
Adding the three number together we get 10 + 6 + 3 = 19.

We attach 9 in 19 as the next digit. But 19 has two digits, so we carry over the 1 in 19. So the last digit from the previous step which is 3 becomes 4 (got by carrying over 3 + 1).

=                      U6 T6
=                            U8 T8
=                                  U3 T3
0    0     0    6     6     7    1           x     386 = 2 5 7 4 9 _ _

Step 2.6

Moving to the next set of digits

Using 6 in 386, the pair product of 7 and 1 is 2 + 0 = 2 (U6T6 = 42 + 06).
Using 8 in 386, the pair product of 1 and _ is 8 + _ = 8 (U8T8 = 08 + –-).
Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –-).
Adding the three number together we get 2 + 8 + _ = 10.

We attach 0 in 10 as the next digit. But 10 has two digits, so we carry over the 1 in 10. So the last digit from the previous step which is 49 becomes 50 (got by carrying over 49 + 1).

=                              U6 T6
=                                   U8 T8
=                                             U3 T3
0    0     0    6     6     7    1                x     386 = 2 5 7 5 0 0 _

Step 2.7

Moving to the next set of digits

Using 6 in 386, the pair product of 1 and _ is 6 + _ = 6 (U6T6 = 06 + –-).
Using 8 in 386, the pair product of _ and _ is _ + _ = _ (U8T8 = –– + –-).
Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –-).
Adding the three number together we get 6 + _ + _ = 6.

6 becomes the last digit of the answer. .

=                                  U6 T6
=                                            U8 T8
=                                                   U3 T3
0    0     0    6     6     7    1                   x     386 = 2 5 7 5 0 0 6

This is your final answer.

Hope this clarifies.

Ah I see. Most of this seem to be arithmetic operations which are covered in the material.

Let me know how it goes.

Wow QianHong.

Where are you from?

Would definitely love to hear more about the things that are tested in eagle flight team club.

Please share your techniques here with the community.

Welcome to the community Jim. How far into the book are you right now?

Hi Toni,

Welcome to the community. Looking forward to seeing your progress updates. Please make sure you complete all the workbooks and post it here.

Abhishek

Ah very useful indeed. If you notice your steps, you are converting the single digit multiplication problem into another multiplication problem 4 x 6 = 24. I would therefore recommend learning the single digit multiplication table by heart.

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