Quarter Wit, Quarter Wisdom Divisibility by Powers of 2

We know the divisibility rules of 2, 4 and 8: For 2 If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2. For 4 If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4. For 8 If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8. A similar rule applies to all powers of 2: For 16 If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16. For 32 If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32. and so on†¦ Let’s figure out why: The generic rule can be written like this: A number M is divisible by 2^n if the last n digits of M are divisible by 2^n. Take, for example, a division by 8 (= 2^3), where  M = 65748048 and  n = 3. Our digits of interest are the last three digits, 048. 48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8. A valid question here is, What about the remaining five digits? Why do we ignore them? Breaking down M, we can see that 65748048 = 65748000 + 048 (weve separated the last three digits). Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000  it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s. All  we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will  also  be divisible by 8. If it is not, 65748048 will not be divisible by 8. In case the last three digits are not divisible by 8, you  can still  find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now   65748000 wont have a  remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8. In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits. We hope you have understood this concept. Let’s take look at a quick  GMAT question to see this  in action: What is the remainder when 1990990900034 is divided by 32 ? (A) 16 (B) 8 (C) 4 (D) 2 (E) 0 Breaking down our given number, 1990990900034 = 1990990900000 + 00034. 1990990900000 ends in five 0s so it is divisible by 32.  34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. Our answer is D. Getting ready to take the GMAT? We have  free online GMAT seminars  running all the time. And, be sure to follow us on  Facebook,  YouTube,  Google+, and  Twitter! Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the  GMAT  for Veritas Prep and regularly participates in content development projects such as this blog!