For instance, say you are given the problem 320∗17{\displaystyle 320*17}. Since 17{\displaystyle 17} is the smaller number, break it into its tens (10){\displaystyle (10)} and ones (7){\displaystyle (7)} components.  This shortcut method works best when the smaller number is between 10 and 19. If the smaller number is between 20 and 99, you'll have to do some extra work to figure out the tens component. As a result, you'll probably find it easier to just do traditional long multiplication. You can also use this method with a 3-digit smaller number as well—in that case, you'll need to break it up into hundreds, tens, and ones. For example, 162 would become 100, 60, and 2. Once again, though, doing standard long multiplication will likely be easier. Now that you've split up the smaller number into tens and ones, use them to create 2 multiplication problems:  320∗10{\displaystyle 320*10} 320∗7{\displaystyle 320*7} Multiplying by 10{\displaystyle 10} is always easy to do. Simply add a zero to the end of the other number. In this case, then, 320∗10=3200{\displaystyle 320*10=3200}. Likewise, when you multiply by 100{\displaystyle 100}, you add 2 zeros, you add 3 zeros when multiplying by 1000{\displaystyle 1000}, and so on. In the current example, the ones problem is 320∗7{\displaystyle 320*7}. Your best bet here is probably to do a fairly straightforward example of long multiplication:  Write down 320{\displaystyle 320}, then write 7{\displaystyle 7} just beneath it, aligned directly below the zero. Draw a 3-digit long line under the 7{\displaystyle 7}. Multiply 7{\displaystyle 7} and each digit of the larger number separately, working from right to left. Since 7∗0=0{\displaystyle 7*0=0}, write a zero under the line, aligned below the 7{\displaystyle 7}. Since 7∗2=14{\displaystyle 7*2=14}, write a 4{\displaystyle 4} just to the left of the zero under the line, and write a small 1{\displaystyle 1} just above the 3{\displaystyle 3} in 320{\displaystyle 320}. This is your reminder to add 1{\displaystyle 1}. Multiply 7∗3=21{\displaystyle 7*3=21}, then add 1{\displaystyle 1} (as indicated by your reminder). Write 22{\displaystyle 22} just to the left of the 3{\displaystyle 3} and zero under the line. Your answer is under the line: 320∗7=2240{\displaystyle 320*7=2240} the tens and ones answers. Right now, you have 320∗10=3200{\displaystyle 320*10=3200} and 320∗7=2240{\displaystyle 320*7=2240}. To get the answer to the original equation, 320∗17{\displaystyle 320*17}, simply add these 2 products together:  Write 2240{\displaystyle 2240} underneath 3200{\displaystyle 3200}, with the right-side zeros aligned. Draw a line underneath 2240{\displaystyle 2240}. Add each column separately, and write the sum below the line:  0+0=0{\displaystyle 0+0=0} 0+4=4{\displaystyle 0+4=4} 2+2=4{\displaystyle 2+2=4} 3+2=5{\displaystyle 3+2=5}   The answer is 5440{\displaystyle 5440}.
प्रॉब्लम के छोटे नंबर को टेन्स और वन्स में बाँट लें: 2 अलग-अलग मल्टिप्लिकेशन प्रॉब्लम्स तैयार करें: पहले बड़े नंबर में जीरो एड करते हुए, टेन्स प्रॉब्लम को सॉल्व करें: वन्स प्रॉब्लम को अलग से सॉल्व करें: टेन्स और वन्स आन्सर को एक-साथ एड करें: