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Difference between revisions of "2023 IOQM/Problem 1"

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==Solution 1(Spacing of squares)==
 
==Solution 1(Spacing of squares)==
  
−
If for any [[integer]] <math>n</math>, if <math>\sqrt{n}</math> is an [[integer]] this means <math>n</math> is a [[perfect square]]. Now the problem reduces to finding the difference between maximum and minimum no. of [[perfect squares]] between <math>4n+1, 4n+2 .... 4n+1000.</math> There are 1000 numbers here.  
+
If for any [[integer]] <math>n</math>, if <math>\sqrt{n}</math> is an [[integer]] this means <math>n</math> is a [[perfect square]]. Now the problem reduces to finding the difference between maximum and minimum no. of [[perfect squares]] in the numbers: <math>4n+1, 4n+2 .... 4n+1000.</math> There are 1000 numbers here.  
  
−
The idea is that for the same range of numbers, the no. of [[perfect squares]] becomes less when the numbers become larger for example, there are 3 [[perfect squares]] between 1 and 10 but none between 50 and 60.  
+
The idea is that for the same range of numbers, the no. of [[perfect squares]] becomes rarer when the numbers become larger.  
  
−
So maximum value of <math>M_n</math> occurs when <math>n</math> is minimum and the minimum value of <math>M_n</math> occurs when <math>n</math> is maximum.  
+
'''For example, there are 3 [[perfect squares]] between 1 and 10 but none between 50 and 60.''' 
  
−
Minimum value of <math>n</math> = 1
+
Of course we will prove this,
 +
 
 +
'''Claim: The distance between 2 consecutive [[perfect squares]] gets larger as they get bigger'''
 +
 
 +
'''Proof:''' Let the 2 consecutive [[perfect squares]] be <math>m^{2}</math> and <math>(m+1)^{2}</math>. Now distance between them (Number of numbers between them) is <math>(m+1)^{2}</math>- <math>m^{2}</math> -1 =<math>2m</math>. So, we notice as <math>m</math> gets larger(i.e. the perfect squares get larger as <math>m^{2}</math> and <math>(m+1)^{2}</math> get larger as m does), <math>2m</math> (distance between <math>m^{2}</math> and <math>(m+1)^{2}</math>)  also does, which means that the distance between the consecutive squares get larger as the consecutive squares get larger (As <math>2m</math> is the measure of distance between them). Hence, this proves our claim.
 +
 
 +
Now, if the distance between 2 perfect squares increases as they get larger, this suffices to prove that perfect squares get rarer
 +
as no.s get larger.
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From this list where perfect squares are highlighted [[RED]] (Except [[1]] which is Blue), it's evident that [[perfect squares]] get rarer as numbers get bigger and bigger.
 +
 
 +
⇒ The maximum value of <math>M_n</math> occurs when <math>n</math> is minimum and the minimum value of <math>M_n</math> occurs when <math>n</math> is maximum.
 +
 +
Minimum value of <math>n</math> = 1 So, the numbers are 5, 6...1004. there are 29 [[perfect squares]] here, so
 +
 
 +
<math>a</math> = <math>max.</math>(<math>M_n</math>)= <math>29</math>
 +
 
 +
Maximum value of <math>n</math> = 1000 So, the numbers are 4001, 4002...5000. there are 7 [[perfect squares]] here, so <math>b</math> = <math>min.</math>(<math>M_n</math>)= <math>7</math>
 +
 
 +
⇒ <math>a-b= 29-7 = \boxed{22}</math>
 +
 
 +
~SANSGANKRSNGUPTA
 +
 
 +
==Solution 2==
 +
All of the sets from <math>1 \leq n \leq 1000</math> will have 1000 elements the most the number of square numbers will be in <math>X_1</math> and least number of perfect squares in <math>X_{1000}</math>. So <math>a=M_1</math> and <math>b=M_{1000}</math>. This is because the gaps between square number is less when n is greater then when n is less. By checking for <math>X_1</math> there would be squares from {<math>3^2, 4^2,..., 31^2</math>}, a total of 29 numbers while in <math>X_{1000}</math> there would be squares from {<math>64^2, 65^2,..., 70^2</math>} a total of 7 numbers, so <math>a=29</math> and <math>b=7</math>, giving us <math>\boxed{a-b=\textbf{22}}</math>
 +
 +
~ Lakshya Pamecha
 +
 
 +
==Video Solutions==
 +
 
 +
Video solution by cheetna: https://www.youtube.com/watch?v=kfEyX5yBdJo
 +
 
 +
Video solution by Unacademy Olympiad Corner: https://www.youtube.com/watch?v=Mm6mXjwU9bY
 +
 
 +
Video solution by Vedantu Olympiad School: https://www.youtube.com/watch?v=4DJXtR4VHEA
 +
 
 +
Video solution by Olympiad Wallah: https://www.youtube.com/watch?v=4HSjmY7d3nA
 +
 
 +
Video solution by : Motion Olympiad Foundation Class 5th - 10th: https://www.youtube.com/watch?v=oVaeHceHXsQ
 +
 
 +
'''Please note that above videos solutions are in Hindi, some in English and some in mixed(Hindi + English).'''
 +
 
 +
~SANSGANKRSNGUPTA
 +
 
 +
==See Also==
 +
[[IOQM]]
 +
 
 +
[[Mathematics competitions]]
 +
 
 +
'''Please note that all problems on this page are copyrighted by THE [https://www.mtai.org.in | MTA(I)]'''

Latest revision as of 21:07, 27 April 2024

Problem

Let $n$ be a positive integer such that $1 \leq n \leq 1000$. Let $M_n$ be the number of integers in the set

$X_n = \left\{\sqrt{4n + 1}, \sqrt{4n + 2}, \ldots, \sqrt{4n + 1000}\right\}$. Let $a = \max\{M_n : 1 \leq n \leq 1000\}$, and $b = \min\{M_n : 1 \leq n \leq 1000\}$.

Find $a - b$.

Solution 1(Spacing of squares)

If for any integer $n$, if $\sqrt{n}$ is an integer this means $n$ is a perfect square. Now the problem reduces to finding the difference between maximum and minimum no. of perfect squares in the numbers: $4n+1, 4n+2 .... 4n+1000.$ There are 1000 numbers here.

The idea is that for the same range of numbers, the no. of perfect squares becomes rarer when the numbers become larger.

For example, there are 3 perfect squares between 1 and 10 but none between 50 and 60.

Of course we will prove this,

Claim: The distance between 2 consecutive perfect squares gets larger as they get bigger

Proof: Let the 2 consecutive perfect squares be $m^{2}$ and $(m+1)^{2}$. Now distance between them (Number of numbers between them) is $(m+1)^{2}$- $m^{2}$ -1 =$2m$. So, we notice as $m$ gets larger(i.e. the perfect squares get larger as $m^{2}$ and $(m+1)^{2}$ get larger as m does), $2m$ (distance between $m^{2}$ and $(m+1)^{2}$) also does, which means that the distance between the consecutive squares get larger as the consecutive squares get larger (As $2m$ is the measure of distance between them). Hence, this proves our claim.

Now, if the distance between 2 perfect squares increases as they get larger, this suffices to prove that perfect squares get rarer as no.s get larger.



1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

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61 62 63 64 65 66 67 68 69 70

71 72 73 74 75 76 77 78 79 80

81 82 83 84 85 86 87 88 89 90

91 92 93 94 95 96 97 98 99 100

101 102 103 104 105 106 107 108 109 110

111 112 113 114 115 116 117 118 119 120

121 122 123 124 125 126 127 128 129 130

131 132 133 134 135 136 137 138 139 140

141 142 143 144 145 146 147 148 149 150

151 152 153 154 155 156 157 158 159 160

161 162 163 164 165 166 167 168 169 170

171 172 173 174 175 176 177 178 179 180

181 182 183 184 185 186 187 188 189 190

191 192 193 194 195 196 197 198 199 200

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261 262 263 264 265 266 267 268 269 270

271 272 273 274 275 276 277 278 279 280

281 282 283 284 285 286 287 288 289 290

291 292 293 294 295 296 297 298 299 300

301 302 303 304 305 306 307 308 309 310

311 312 313 314 315 316 317 318 319 320

321 322 323 324 325 326 327 328 329 330

331 332 333 334 335 336 337 338 339 340

341 342 343 344 345 346 347 348 349 350

351 352 353 354 355 356 357 358 359 360

361 362 363 364 365 366 367 368 369 370

371 372 373 374 375 376 377 378 379 380

381 382 383 384 385 386 387 388 389 390

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531 532 533 534 535 536 537 538 539 540

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571 572 573 574 575 576 577 578 579 580

581 582 583 584 585 586 587 588 589 590

591 592 593 594 595 596 597 598 599 600

601 602 603 604 605 606 607 608 609 610

611 612 613 614 615 616 617 618 619 620

621 622 623 624 625 626 627 628 629 630

631 632 633 634 635 636 637 638 639 640

641 642 643 644 645 646 647 648 649 650

651 652 653 654 655 656 657 658 659 660

661 662 663 664 665 666 667 668 669 670

671 672 673 674 675 676 677 678 679 680

681 682 683 684 685 686 687 688 689 690

691 692 693 694 695 696 697 698 699 700

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711 712 713 714 715 716 717 718 719 720

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741 742 743 744 745 746 747 748 749 750

751 752 753 754 755 756 757 758 759 760

761 762 763 764 765 766 767 768 769 770

771 772 773 774 775 776 777 778 779 780

781 782 783 784 785 786 787 788 789 790

791 792 793 794 795 796 797 798 799 800

801 802 803 804 805 806 807 808 809 810

811 812 813 814 815 816 817 818 819 820

821 822 823 824 825 826 827 828 829 830

831 832 833 834 835 836 837 838 839 840

841 842 843 844 845 846 847 848 849 850

851 852 853 854 855 856 857 858 859 860

861 862 863 864 865 866 867 868 869 870

871 872 873 874 875 876 877 878 879 880

881 882 883 884 885 886 887 888 889 890

891 892 893 894 895 896 897 898 899 900

901 902 903 904 905 906 907 908 909 910

911 912 913 914 915 916 917 918 919 920

921 922 923 924 925 926 927 928 929 930

931 932 933 934 935 936 937 938 939 940

941 942 943 944 945 946 947 948 949 950

951 952 953 954 955 956 957 958 959 960

961 962 963 964 965 966 967 968 969 970

971 972 973 974 975 976 977 978 979 980

981 982 983 984 985 986 987 988 989 990

991 992 993 994 995 996 997 998 999 1000

From this list where perfect squares are highlighted RED (Except 1 which is Blue), it's evident that perfect squares get rarer as numbers get bigger and bigger.

⇒ The maximum value of $M_n$ occurs when $n$ is minimum and the minimum value of $M_n$ occurs when $n$ is maximum.

Minimum value of $n$ = 1 So, the numbers are 5, 6...1004. there are 29 perfect squares here, so

$a$ = $max.$($M_n$)= $29$

Maximum value of $n$ = 1000 So, the numbers are 4001, 4002...5000. there are 7 perfect squares here, so $b$ = $min.$($M_n$)= $7$

⇒ $a-b= 29-7 = \boxed{22}$

~SANSGANKRSNGUPTA

Solution 2

All of the sets from $1 \leq n \leq 1000$ will have 1000 elements the most the number of square numbers will be in $X_1$ and least number of perfect squares in $X_{1000}$. So $a=M_1$ and $b=M_{1000}$. This is because the gaps between square number is less when n is greater then when n is less. By checking for $X_1$ there would be squares from {$3^2, 4^2,..., 31^2$}, a total of 29 numbers while in $X_{1000}$ there would be squares from {$64^2, 65^2,..., 70^2$} a total of 7 numbers, so $a=29$ and $b=7$, giving us $\boxed{a-b=\textbf{22}}$

~ Lakshya Pamecha

Video Solutions

Video solution by cheetna: https://www.youtube.com/watch?v=kfEyX5yBdJo

Video solution by Unacademy Olympiad Corner: https://www.youtube.com/watch?v=Mm6mXjwU9bY

Video solution by Vedantu Olympiad School: https://www.youtube.com/watch?v=4DJXtR4VHEA

Video solution by Olympiad Wallah: https://www.youtube.com/watch?v=4HSjmY7d3nA

Video solution by : Motion Olympiad Foundation Class 5th - 10th: https://www.youtube.com/watch?v=oVaeHceHXsQ

Please note that above videos solutions are in Hindi, some in English and some in mixed(Hindi + English).

~SANSGANKRSNGUPTA

See Also

IOQM

Mathematics competitions

Please note that all problems on this page are copyrighted by THE | MTA(I)

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