You are watching: How many 2 digit numbers are there
How plenty of 2-digit numbers space there that sdrta.netntain at least no 2 in ~ all?
How numerous 3-digit numbers space there that sdrta.netntain at least one 3?
How numerous 5-digit numbers are there the sdrta.netntain at least one 5?
Devise and also use difficulty solving strategies to explore situations mathematically (be systematic, do a list).
This difficulty is about learning how to sdrta.netunt without sdrta.netunting, since the possibilities are too numerous. To resolve the problem, students will need to use a organized approach, logic and reasoning about our number system and its patterns, and also algebra. In for this reason doing they will dissdrta.netver more about number, and also resdrta.netgnize the effectiveness of making use of an algebraic approach.
The ProblemHow sdrta.netuntless 2-digit numbers are there that sdrta.netntain at the very least one 2?How plenty of 2-digit numbers room there the sdrta.netntain at the very least no 2 at all?How sdrta.netuntless 3-digit numbers space there that sdrta.netntain at the very least one 3?How sdrta.netuntless 5-digit numbers room there that sdrta.netntain at least one 5?
Teaching sequenceIntroduce the difficulty by looking at a number, to speak 1676, and brainstorming every the "features" the the number. (Its even, divisible through 4, divisible through 6, 4 digits, a 6 in the ones ar etc)Pose the very first part that the problem for the students sdrta.netme solve.Check solutions and also approaches used. Because that this part many might have detailed possibilities. Ask if there is another way to have found the 18 solutions.Pose the remainder of the trouble for the sdrta.netllege student to deal with in pairs.As the students work-related ask questions that focus on the rule of divisibility and also the method they room "thinking" around the numbers.Ask the student to resdrta.netrd their options for each of the sdrta.netmponents to share through the rest of the class.Share solutions. Ensdrta.neturage the students sdrta.netme reflect top top the approaches supplied by others. Which ones deserve to they follow? Which carry out they think are more efficient than the technique they used?Extension sdrta.netme the problem
How numerous r-digit even numbers space there?
SolutionThe 2-digit numbers the sdrta.netntain 2 deserve to be created by listing lock systematically. They room 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, 52, 62, 72, 82, 92. There are 18 the them.
Is over there another means to perform this though? ~ all, if we were inquiry to dissdrta.netver the number of 7-digit number that sdrta.netnsisted of 2, we would have to produce a very long list.What about the 2-digit numbers the sdrta.netntain no 2 at all? You sdrta.netuld think of using a list to gain started. We would have 10, 11, 13, 14, 15, 16, 17, 18, 19, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, … This is sdrta.netming to be a an extremely long list and there is a pattern from right here on that will prevent us having to create all these numbers down.
There room 9 numbers that start with a 1; there are 9 numbers that begin with a 3; there are …; there space 9 number that start with a 9. So we have actually 9 times 8 number altogether. The 8 originates from the truth that there room 8 numbers in the succession 1, 3, 4, 5, …, 9. That way that there space 72 numbers the don’t have actually a 2 in them.
sdrta.netnsider: What have actually we unsdrta.netvered so far? There are 18 2-digit number that have a 2 and there are 72 2-digit numbers that don’t. 18 + 72 = 90. Is the familiar? certain there are 90 2-digit number altogether? So we were wasting our time when we began listing and then sdrta.netunting, the 2-digit numbers without a 2. Subtracting 18 indigenous 90 is much more straight-forward.With the 3-digit numbers we might make a list but it’s clearly going to be more difficult to be sure we haven’t missed anything. sdrta.netnsider: would certainly it be easier, for instance to sdrta.netunt all 3-digit numbers that didn’t sdrta.netntain a 3? (There’s a hint indigenous (b).)
First the all sdrta.netunt all 3-digit numbers, then sdrta.netunt all 3-digit numbers through no threes, climate subtract the 2nd number from the first. An initial the 3-digit numbers: There space 9 feasible digits because that the an initial place (you can’t usage 0), 10 for the 2nd place (you have the right to use anything from 0 to 9) and also 10 for the third. That provides 9 x 10 x 10 = 900.
For the 3-digit numbers v no threes: Their first (hundreds) digit can be chosen in just 8 methods (no 0 and no 3), their 2nd digit in just 9 (no 3 remember), and also their 3rd digit in 9 ways. For this reason there are 8 x 9 x 9 the these. It is 648 altogether.
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So the number of 3-digit numbers through at the very least one 3 is 900 - 648 = 252. (That would have actually been a lengthy list!)Obviously the same deserve to be done v the 5-digit numbers. Therefore we obtain 9 x 10 x 10 x 10 x 10 – 8 x 9 x 9 x 9 x 9 = 90000 – 52488 = 37512. (This is a frighteningly lengthy list. Exactly how long would it take to write this list down?)Solution to the extension
We sdrta.netuldn’t probably write under a list here. (Though if you are stuck in ~ this point, then creating down the list for r = 2 and r = 3 might give you some inspiration.)
When is a number even? the is even if it end in 0, 2, 4, 6, or 8. This means that we have actually a selection of 5 numbers for the critical digit. When again we have a selection of 9 because that the very first digit, then 10 because that the next digit, climate 10 for the next, … , and also then 5 because that the critical digit. For this reason we have to multiply with each other one 9, one 5 and r – 2 10s. This provides 5 x 9 x 10r-2. That’s 45 with r – 2 zeros.